a ball is thrown straight upward and returns to the thrower's hand after 3.00s in the air. A second ball is thrown at an angle of 30.0 deg with the horizontal. At what speed must the second ball be thrown so that it reached the same height as the one thrown vertically?

Answer 1

Use kinematics to calculate the height for the firstball,

which is Vf=0

a=9.8

t=1.5
because you take the total time it take to go up and come back down and divide it by 2. 3/2 = 1.5



getting an initial velocity of
Voy= 14.7m/s
V = t(9.8)
= 1.5(9.8)
= 14.7


This must have the same velocity in the y to get that height of 11.025m calculated in part 1, so just use trig to get the answer.


14.7/sin30
=29.4 m/s