Question

A ball is tossed upward and returns to its original position after 11.9 seconds. If another ball is thrown at an angle of 44.8 degrees above the horizontal, at what speed, in m/s, must it be thrown such that it returns to the same height at the same time as the first ball?

Answer 1

@dude

Answer 2

For the first ball, apply the equation of motion
\ 0=u∗11.9−1/2∗9.8∗11.92 The initial velocity of the first ball is =58.31 m/s Let the velocity of the second ball be v m/s Then \[vsin(44.8º)=u\] \[v=58.31/\sin(44.8º)= 82.8 m/s\]