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3 years ago

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A ball is projected horizontally from the top of a bertical building 25.0m above the ground level with an initial velocity of 8.

25m/s . Calculate :
a. How long is the ball in flight before striking the ground
b. How far from the building does the ball strike the ground

1 answer:

kirill115 [55]3 years ago

7 0

Answer:

Solution given:

height [H]=25m

initial velocity [u]=8.25m/s

g=9.8m/s

now;

a. How long is the ball in flight before striking the ground?

Time of flight =?

Now

Time of flight= $\sqrt{\frac{2H}{g}}$

substituting value

= $\sqrt{\frac{2*25}{9.8}}$
=2.26seconds

<h3><u>the ball is in flight before striking the ground for 2.26seconds</u>.</h3>

b. How far from the building does the ball strike the ground?

<u>H</u><u>o</u><u>r</u><u>i</u><u>z</u><u>o</u><u>n</u><u>t</u><u>a</u><u>l</u><u> </u>range=?

we have

Horizontal range=u* $\sqrt{\frac{2H}{g}}$

=8.25*2.26
=18.63m

<h3><u>The ball strikes 18.63m far from building</u>. </h3>

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