A 0.250kg ball is thrown with

Question

an initial speed of 35.0 m/s at a 40.0° angle above the horizontal. It is thrown from the top of a building which is 17.0m above the ground. Ignore air resistance.

What is the maximum height above the ground the ball reaches?

Answer 1

x = 35cos40°t; v_x = 35cos40°;

y = 17 + 35sin40°t - 4.9t²; v_y = 35sin40° - 9.8t.

At maximum height v_y = 0, 35sin40° - 9.8t = 0, t = 35sin40°/9.8 = 2.296 s.

So maximum height h_max = y(2.296) = 17 + 35sin40°·2.296) - 4.9(2.296)² = 42.82 m