Projectile Motion

Question

A ball is thrown straight upward and returns to the thrower’s hand after 3.7 s in the air. A second ball is thrown at an angle of 45◦ with the horizontal. At what speed must the second ball be thrown so that it reaches the same maximum height as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

Answer 1

Ball 1) Total t = 3.7s, so time to reach maximum height is 1.85s. From v = v_o + at, and v = 0 at max height, 0 = v_o -(9.8)(1.85), v_o = 18.13 m/s

Ball 2 needs the same v_y component as ball 1, but now v_y = v_o sin 45, total v_o = (18.13)/(0.7071) = 26 m/s [two sig figs].