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3 years ago

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A ball is thrown horizontally from the top of a building 21.8 m high. The ball strikes the ground at a point 101 m from the base

of the building. The acceleration of gravity is 9.8 m/s 2 . Find the time the ball is in motion. Answer in units of s. 008 (part 2 of 4) 10.0 points Find the initial velocity of the ball. Answer in units of m/s.

1 answer:

riadik2000 [5.3K]3 years ago

6 0

Answer:

t=2.10 s

u= 47.40 m/s

Explanation:

given that

h= 21.8 m

x= 101 m

g=9.8 m/s²

Lets take horizontal speed of ball = u m/s

The vertical speed of the car at initial condition is zero ( v= 0).

We know that

$h=vt+\dfrac{1}{2}gt^2$

v= 0 m/s

$h=\dfrac{1}{2}gt^2$

now by putting the values

21.8 = 1/2 x 9.8 x t²

t=2.10 s

This is time when ball was in motion.

Now in horizontal direction

x = u .t

101 = u x 2.1

u= 47.40 m/s

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