Question

A ball is thrown vertically up with a certain velocity from the top of a tower of height 40m. At 4.5m above the top of the tower its speed is exactly half of that it will have at 4.5 m below the top of the tower. Find the maximum height reached by the ball above the ground? ( Ans : 47.5 m)

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Answer 1

Dear student,

● Answer -

h = 47.5 m

◆ Explanation -

Let u be the initial velocity of the ball with which it was thrown.

When the ball is thrown upwards -

v1^2 = u^2 + 2as

v1^2 = u^2 + 2(-10)(4.5)

v1^2 = u^2 - 90

When the ball is thrown downwards -

v2^2 = u^2 + 2as

v2^2 = u^2 + 2(10)(4.5)

v2^2 = u^2 + 90

But given that v2 = 2v1,

(2v1)^2 = u^2 + 90

4 (u^2 - 90) = u^2 + 90

4u^2 - u^2 = 90 + 360

u^2 = 450/3

u^2 = 150

Height reach by the ball above tower -

v^2 = u^2 + 2as

0^2 = 150 + 2(-10) × s

s = 150/20

s = 7.5 m

Maximum height reached by the ball above ground will be -

h = 40 + 7.5

h = 47.5 m

Thanks for asking...

Answer 2

Answer:

e

Explanation: