1.- a ball is hurled straight up at a speed of 15.0m/s, leaving the hand of the

Question

1.- a ball is hurled straight up at a speed of 15.0m/s, leaving the hand of the thrower 3.00 m above ground. (a) compute the times and the ball's speed when it passes an observer sitting at a window in line with the trow 10.0 meters above the point of release. (b) What is the total distance traveled by the ball just as it passes the window the second time?

2.- Michael drops a penny into a river from a bridge 44 m above the water. Another penny is thrown vertically down by his friend Jeremy 1.00 second after . Both stones strike the water at the same time. What is the initial speed of the second stone

Explanation / Answer

1. initial velocity of the ball is u= 15m/s the ball leaves the hands from 3m above the ground height. a) for a observer at 10m height. h = 10 metres. according to newtons law of motion we know that s =ut + 0.5 at*t now s=10 m and a =-9.8 m /(s*s) so by solving we get 10= 15*t + (-9.8) t*t therefore time =0.765 seconds . for computing the speed.. we know that v=u +at so wen u=15m/s and a=-9.8m/(s*s) and t=0.765 seconds v = 9.264 m/s . b) total distance travelled by the ball when its velocity reached 0 is H so by v*v=u*u +2a*s we get 0= 15*15 + 2(-9.8) H H =11.47 metres. the window is at 10 metres height ..so the distance travelled by the ball after crossing the window = 1. 47 metres total distance travelled = 1.47 *2 = 2.94 metres. 2) time taken by the first stone to reach be t seconds. so s= u*t +0.55 * a *t*t 44= 0+ 0.5*9.8*t*t. t=2.99 = 3 seconds time taken by second stone = 2 seconds s= u*t +0.55 * a *t*t 44= u*2 + 4.9 *2*2 therefore u =12.2 m/(s*s)

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