Question

a ball is projected vertically upwards, if it's velocity after reaching half of its max height is 20m/s the find the height of the tower​

Answer 1

Answer:

at half the max height , vel = 20m/s

h = u^2/2g

u = √2gh

v^2 - u^2 = 2as

(20)^2 - 2gh = 2(-g)(h/2)

400-20h = -10h

10h = 400

h = 40 metres

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Answer 2

Correct question:

A ball is projected vertically upwards, if it's velocity after reaching half of its maximum height is 20m/s. Find the maximum height reached by the ball.

Answer:

Maximum height reached by ball

= 40 m

Step by step explanations :

Given that,

A ball is projected vertically upward

and it's velocity after reaching half of its maximum height is 20m/s

let the height of the tower be h

so,

velocity of ball will 20 m/s at h/2

Let the initial velocity of the ball at h/2 be u

now,

we have,

Initial velocity(u) = 20 m/s

final velocity(v) = 0 m/s

[ball will at rest at its maximum height]

gravitational acceleration(g) = -10 m/s²

height = h/2

by the gravitational equation of motion,

v² = u² + 2gh

putting the values,

0² = 20² + 2(-10)h/2

0 = 400 - 10h

-10h = -400

h = -400/-10

h = 400

so,

Maximum height reached by ball

Maximum height reached by ball = 40 m