Physics question
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A boy at the bottom of a 150m high cliff throws a tennis ball vertically upwards. A boy at the very top of the cliff drops a golf ball so that it hits the tennis ball while bot hballs are still in the air. The acceleration of both balls can be assumed to be 10ms^-2 downwards.
a. With what speed is the tennis ball thrown so that the golf ball strikes it at the maximum height of the path?
b. What is the position of the tennis ball when the golf ball strikes it? -
let t be time of "collision" in seconds
distance travelled by golf ball = 1/2 x t/s x t(10)m/s
= 5t^2 metres (pythagoras theorem)
principle of conversation of energy -> can assume that at maximum height,
kinetic energy = 0
tennis ball-
G.P.E at maximum height = K.E. at bottom of cliff
mgh = 1/2 mv^2
20h = v^2
h = (v^2)/20
deceleration of pitched tennis ball = 10 m/s^2
t(10) = v
100t^2 = v^2
150m = 5t^2 + (100t^2)/ 20
t = root of 15.
initial speed of ball = 10 x (root of 15)
=38.7 m/s
position of tennis ball = (v^2)/20
= 75m
erm. im not sure if this is correct, never done such a question before. i'm sec 4. is this in my syllabus, lol? anyway i tried subbing in the values, does seem to be correct but i made a few assumptions i'm not sure if its correct -
a. u = sqrt(15)*10 = 38.7 m/s
b. x = 75m -
didnt get some of your workings. mind explaining? and tamago, no workings?Originally posted by kristovorus:let t be time of "collision" in seconds
distance travelled by golf ball = 1/2 x t/s x t(10)m/s
= 5t^2 metres (pythagoras theorem)
principle of conversation of energy -> can assume that at maximum height,
kinetic energy = 0
tennis ball-
G.P.E at maximum height = K.E. at bottom of cliff
mgh = 1/2 mv^2
20h = v^2
h = (v^2)/20
deceleration of pitched tennis ball = 10 m/s^2
t(10) = v
100t^2 = v^2
150m = 5t^2 + (100t^2)/ 20
t = root of 15.
initial speed of ball = 10 x (root of 15)
=38.7 m/s
position of tennis ball = (v^2)/20
= 75m
erm. im not sure if this is correct, never done such a question before. i'm sec 4. is this in my syllabus, lol? anyway i tried subbing in the values, does seem to be correct but i made a few assumptions i'm not sure if its correct -
erm. i treated it more like a maths question rather than physics. which part u didnt quite understand? add me on msn lor. i try answer. msn on my profile
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its okay i understand your working on further looking through. and also asked on yahoo answersOriginally posted by kristovorus:erm. i treated it more like a maths question rather than physics. which part u didnt quite understand? add me on msn lor. i try answer. msn on my profile
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For tennis ball,Originally posted by Deadly:A boy at the bottom of a 150m high cliff throws a tennis ball vertically upwards. A boy at the very top of the cliff drops a golf ball so that it hits the tennis ball while bot hballs are still in the air. The acceleration of both balls can be assumed to be 10ms^-2 downwards.
a. With what speed is the tennis ball thrown so that the golf ball strikes it at the maximum height of the path?
b. What is the position of the tennis ball when the golf ball strikes it?
the boy throws it upwards.
s = ut + 1/2 at^2
a = -10 if I am taking upward as positive.
s = ut - 5t^2
If you have drawn the displacement vs time graph of the tennis ball, it will be a
curve, or frowning curve. The maximum height of the path of tennis ball implies the stationary point of the curve in this case.
Hence,
ds/dt = u - 10t
when ds/dt = 0,
u - 10t = 0
u = 10t where u represents the initial speed of tennis ball.
sub u =10t into s = ut - 5t^2
s = 10t^2 - 5t^2
= 5t^2
This means that at v = 0 which is the maximum height, displacement travelled by the ball upwards = 5t^2.
For golf ball,
the boy drops it.
u = 0
a = 10
s = ut + 1/2 at^2
In this case, s is the displacement travelled by the golf ball taking downwards to be positive.
s = 1/2(10)(t^2)
= 5t^2
This means that at time t seconds, the golf ball will travel 5t^2 downwards.
Since the balls clash at this point, assuming that the diameter of the ball is negligible,
5t^2 + 5t^2 = 150
10t^2 = 150
t = root(15)
Initial speed of tennis ball = 10t
= 10root(15)
Position of tennis ball = 5t^2
= 5(15)
= 75 m -
paiseh. was actually trying to confirm his answer is correct. same working.Originally posted by Deadly:and tamago, no workings?
either u approach it as differentiation qn, find maxima which i dun really recommend since it's physics, but here i used it.
Let u be the initial speed.
d/dt(ut-5t²) = u-10t
when (ut-5t²) is max, u-10t=0 and t=u/10
150-5t²=ut-5t²
150=u(u/10)=u²/10
u²=1500
u=10*sqrt(15) -
but if you use common sense... it tells u that the faster u throw the tennis ball upwards, the further it will travel before it will hit the golf ball right? so there is no real specific speed it must travel in order to the reach the maximum height.
anyone wanna enlighten me? -
I also don't really understand your qn at first until something enlightens me.Originally posted by Deadly:but if you use common sense... it tells u that the faster u throw the tennis ball upwards, the further it will travel before it will hit the golf ball right? so there is no real specific speed it must travel in order to the reach the maximum height.
anyone wanna enlighten me?
Maximum height in this qn refers to maximum height of the tennis ball.
Imagine, while you throw a tennis ball upwards, another guy drops a golf ball.
If you apply a high initial spped just like what you have said earlier, the golf ball will not hit the tennis ball just as it reaches maximum height. The golf ball will hit the tennis ball at time t when the tennis ball is still slowing down. -
Originally posted by tanjun:
what u mean the golf ball wont hit the tennis ball as it reaches maximum height?
If you apply a high initial spped just like what you have said earlier, the golf ball will not hit the tennis ball just as it reaches maximum height. The golf ball will hit the tennis ball at time t when the tennis ball is still slowing down.
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Let me clarify myself a bit.Originally posted by Deadly:what u mean the golf ball wont hit the tennis ball as it reaches maximum height?
Imagine if you throw a tennis ball upwards with a high initial speed, while another person drops a golf ball. The tennis ball will be zooming upward, slowing down along the way. The maximum height of the ball with this initial speed may be 140 m for example after a certain time t. However, at this point, the golf ball may have already travelled 30 m at the same time t and it is at the 120m mark.
Thus, it is this precise time t whereby the distance travelled by the golf ball is equal to the maximum upward distance travelled by the tennis ball. We want to find this time t. -
The question asked "With what speed is the tennis ball thrown so that the golf ball strikes it at the maximum height of the path? Guess that answers your question.Originally posted by Deadly:but if you use common sense... it tells u that the faster u throw the tennis ball upwards, the further it will travel before it will hit the golf ball right? so there is no real specific speed it must travel in order to the reach the maximum height.
anyone wanna enlighten me?