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A numerical problem from motion in 1 Dimension

A boy plays with a ball on a field surrounded by a fence which has a height of 2.5m . He kicks the ball vertically up from a height of 0.4m with a speed of 14 ms−1. (a) What is the maximum height of the ball above the fence? (b) What is the time taken to reach the maximum height? (c) How long is the ball above the height of the fence? (d) What is the velocity of the falling ball at the height of the fence? (e) What the acceleration of the ball after 1s ?

Asked John

Answer:

(a)

u=14 m/s

v=0 m/s at max height

y0=0.4 m

y=?

0-(14×14)=2(-9.8)x(y-.o4)

196=19.6x(y-0.4)

y=10.4m

The height above fence = 10.4-2.5= 7.9m

(b)

h=y-y0=10m

u=14 m/s

v=0 m/s

a=-9.8 m/s^2

v=u+at

t=(v-u)/a = (0 – 14)/(-9.8) = 1.43 s

(c)

Will be updated soon

(d) -9.8 m/s^2 (Acceleration is equal to the acceleration due to gravity, which is constant throughout the motion)

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