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home / study / science / physics / questions and answers / a small rock is projected vertically upwards with ... Question a small rock is projected vertically upwards with a speed of 10.0 M/s from the side of a bridge at a point 65.0 M above the water below. g=9.8 m/s2. a) how much later (t final) will the stone strike the water surface? b) find the velocity (V final) of the rock just before it hits the water. c) how high (h) will it rise above the bridge? d) how long does it take (t peak, time at peak) for the stone to rise to the peak of it flight ? e) what is the magnitude and direction of the stone's velocity (V Peak) and acceleration (a p) at the instant when it is atthe peak of it's flight   

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Explanation / Answer

Taking downwards as positive and upwards as negative.

g = 9.8 m/s2

initial velocity u = - 10 m/s

At the peak the velocity will be zero. (because, if at all it has a little velocity left, it will go little higher and the current point wouldn't be the peak)

so for v = 0 , we have

v = u +at

0 = -10 + gt

t = 10 / g = 10 / 9.8 = 1.020408163 seconds to reach the peak (answer to part d)

using the formula v2 - u2 = 2gh (where h is the peak height from the bridge , not the ground)

02 - ( -10 )2 = 2gh

-100 / 2g = h

h = - 5.102040816 m (answer to part c)

(we got a negative value just because of our convention, we have taken upwards as negative and this height is measured from the bridge upwards)

From this top point till it hits the water, i.e in the return trajectory, the distance the stone need to cover to hit the water is h + 65 = 5.102040816 + 65 = 70.10204082 m (Here we have taken h positive because, in the return path, we are measuring the distance from the peak point in the downward direction)

so the initial velocity of the return trajectory, that is the velocity at the peak point is u' = 0

and the acceleration is always g = 9.8 m/s2   downwards (answer to part e)

let the final velocity of this return trajectory be v', which is the velocity just before the stone hits the water

So, v'2 - u'2 = 2gs

v'2 - 02 = 2 x 9.8 x 70.10204082

v' = 37.06750599 m/s (answer to part b)

time taken for this return path be T

v' = u' + gT

T = (v' - u' )/g

T = 37.06750599/9.8

T = 3.78239857 seconds

So, the entire time taken to reach the peak and then to fall and hit the water is t + T

t + T = 1.020408163 + 3.78239857 = 4.802806733 seconds (answer to part a)

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