A ball is projected with an initial velocity of 40 meter per second and reached maximum height of 160 meters calculate tge angle of projection

There's a problem with the question as given. Even with a maximum projection angle of θ = 90°, the initial velocity is not large enough to get the ball up in the air 160 m. With angle 90°, the ball's height y at time t would be

y = (40 m/s) t - 1/2 g t ²

Set y = 160 m, and you'll find that there is no (real) solution for t, so the ball never attains the given maximum height.

From another perspective: recall that

v ² - v₀² = 2a ∆y

where

• v₀ = initial velocity

• v = final velocity

• a = acceleration

• ∆y = displacement

At its maximum height, the ball has zero vertical velocity, and ∆y = maximum height = 160 m. The ball is in free fall once it's launched, so a = -g.

So we have

0² - (40 m/s)² = -2g (160 m)

but this reduces to

(40 m/s)² = 2 (9.8 m/s²) (160 m)

1600 m²/s² ≠ 3136 m²/s²

2.4086 * 10^(-4)  m^3

Explanation:

Volume of sphere = 4/3 * pi * r1^3

Volume of cylinder = pi*r2^2*h

r1 = 0.05 m

r2 = 0.03 m

Hence,

Resulting Volume = Volume of sphere - Volume of cylinder