Question

From a building of height 80 m, a ball is projected with a velocity of 20 m/s. At
what distance from the bottom of the building will the ball strike the ground? Also,
what will be the magnitude and direction of velocity of the ball one second after its
projection from the top of the building?
(Take g = 10 m/s2.)​

Answer 1

To find the distance from the bottom of the building where the ball strikes the ground, only horizontal motion is considered. So,

u = 20 cosθ ms⁻¹

a = 0 ms⁻²,  because no horizontal force is acting on the ball.

So, v = u + at = 20 cosθ ms⁻¹.

Here, θ is the angle of projection of the ball. We have to find 's'.

    s = ut + (at²)/2

⇒  s = 20 cosθ · t  m

After t = 1 second, the velocity will have the magnitude,

$\sqrt{u^2+g^2t^2-2ugt\sin\theta}=\sqrt{400+100-2\cdot20\cdot10\sin\theta}\\\\=\sqrt{500-400\sin\theta}$

and will make an angle,

$\tan^{-1}\left(\dfrac{u\sin\theta-gt}{u\cos\theta}\right)=\tan^{-1}\left(\tan\theta-\dfrac{gt}{u\cos\theta}\right)$

with the x - axis.

If the angle θ was known, we could have a unique value for these, and we could find the value of 't', time taken by the ball to reach the ground. But remember, here it's considered that the ball is projected "upwards" from the top of the building.