Question

A ball is thrown with an initial velocity of 100m/s at an angle of 30${}^0$ above the horizontal . How far from the throwing point will the ball attain its original level? Solve the problem without using formula for horizontal range

Answer 1

Given,

Initial velocity $=100m/s$

Angle $={30}^0$

So,

The initial velocity makes an angle 30° with horizontal. So,its vertical component will be $100sin{30}^0=50m/s.$

The time taken by the ball to acheive max height will be at a point when its vertical velocity will be zero. So $Vsinx=at$

$T=\frac{vsinx}{a}=5seconds$.

So after 5 seconds it will reach its max height.

Following the symmetry,it will land after 10seconds.

So the total horizontal distance covered will be $vcosx\times 10\approx 866m$