Question

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Answer 1

Hey there!

So, The ball is thrown up with a velocity of 25m/s , and the motion of the ball is influenced by gravity. Hence, The ball is in motion under gravity.

As the ball is moving up ( away from Earth's centre) , The acceleration is -g.

Given Velocity it is thrown is it's initial velocity which is equal to 25m/s , When it reaches the maximum height its velocity becomes 0

We need to find the maximum height it reaches and time taken to reach

Putting all the information again,
v = 0m/s
u = 25m/s
a = -g = -9.8m/s²

From the first equation of motion,
v = u + at
0 = 25 - 9.8t
25 = 9.8t
t = 25/9.8 = 2.55seconds

We know that, From the third equation of motion
v² - u² = 2as
0² - u² = 2(-g)H
u² = 2gH
25² = 2(9.8 ) H
625 / 19.6 = H
H = 31.88

You can get more easy values if you consider g = 10m/s²
If you consider like that, t = 2.5s , H = 31.25m .

Therefore, The maximum height the body will reach is 31.25m , and time taken to reach it is 2.5s