Question

A ball is thrown up at a speed of 4.0m/s .find the maximum height reached by the ball.take g=10 m/s^2

Answer 1

here is your answer

given :

initial velocity (u) = 4 m/s

final velocity (v) = 0

acceleration due to gravity (g) = 10 m/s^2

height = ?

solution :

by using 3rd equation of motion

v^2 = u^2-2gh

0 = (4)^2-2*10*h

20h = 16

h = 4/5 or 0.8 meter

so, the maximum height of the ball will be 0.8 meter.

Answer 2

hii friend. ...

here is ur answer...
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Solution:-
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Let us take vertically upward direction as the positive y-axis .

we have u=4.0 m/s

a= -10m/s^2

at the highest point the velocity becomes zero .....

using the formula. ...

v^2=u^2+2ay

0=(4.0)^2+2 (-10).y

y=16/20=0.80 m.....
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hope it will help you. ....

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BY silu12@♡♡♡♡♡♡♡♡♡

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