Question

A ball is thrown straight upward and returns to the thrower's hand after 3.50 s in the air. A second ball is thrown at an angle of 43.0° with the horizontal. At what speed must the second ball be thrown so that it reaches the same height as the one thrown vertically?

Answer 1

Suppose time taken by the first ball for upward is t

So 2t=3.5 s

t=1.75 s

Using equation of the motion

here V= 0 (at maximum height ) g=9.8 m/s , t=1.75 s

Again using equation of the motion

for second ball, It is projectile motion

We know that   

Answer