A ball is thrown straight upward and returns to the thrower's hand after 2.1 s in the air. A second ball is thrown at an angle of 41° with the horizontal.

Question

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/s². Answer in m/s.

Answer 1

The height y of Ball 1 at time t is y = v₁t – gt ²/2, so its maximum is when dy/dt = v₁ – gt → 0 when t → v₁/g, and it returns to y → 0 at t → 2v₁/g = T.

The height of Ball 2 is v₂t sinθ – gt ²/2 with maximum when t → v₂/g sinθ.

Equating the 2 maximum heights gives v₁(v₁/g) – g(v₁/g)²/2 = v₂(v₂/g sinθ)sinθ – g(v₂/g sinθ)²/2,

whence v₂ = v₁cscθ = gT/2 cscθ.

Answer 2

time for a ball thrown straight up is 2v₀/g derived from v = v₀ + at with v = -v₀, t given, and a = -g

Rearranging, you can solve for v₀ = (1/2) gt

In order to reach the same maximum height in the 2d problem, just make v_{0,part 1} = v_o,y as the y component of velocity will determine the max height.

v_0,Part2 = v_0,Part1/sin(θ₀)

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