A ball is thrown straight upward and returns to the thrower's hand after 3s in the air.?

Question

A second ball is thrown at an angle of 30 degrees with the horizontal. The acceleration of gravity is 9.8m/s/s. At what speed must the second ball be thrown so that is reaches the same height as the one thrown vertically? Answer in units of m/s.

Answer 1

The speed v with which a ball goes straight up or comes back down is related to the total time it takes by

- v = v - g t, so that t = 2v / g, or

v = gt/2 = 9.8*3/2 m/s = 14.7 m/s.

The height h reached is given by h = v^2 / 2g.

The second ball is thrown up with speed V at an angle of 30 deg. above the horizontal. As only the UPWARD COMPONENT of the speed matters, the height then reached is

V^2 (sin 30)^2 / 2g.

So to reach the same height, you'd need V = v / sin(30)

= 2v = 29.4m/s.

Live long and prosper.

Answer 2

u = gt; where t = T/2 = 1.5 sec, the time to reach v = 0 velocity at the apex of the trajectory. u is the launch velocity ~ 10*1.5 = 15 m/sec. [I'm using g = 10 m/sec^2 to show how this is worked, you can use 9.8 if you like.]

Thus, u = U sin(30) where U is the launch velocity of the second ball; so that U = u/sin(30) = 2*15 = 30 m/sec. In other words, the vertical component of U has to equal u found earlier so that the ball will reach the same height as the first ball. This results, because the vertical kinetic energies of both balls must be the same to reach the same potential energy at the same height h (PE = mgh, we assume the mass of the two balls is the same).