Question

A stone is thrown with the velocity of 10 ms^-1 from the tower at an angle of 30° upwards with horizontal. It reaches the ground after 5 s. Calculate the height of the tower and range of the stone, (g = 10ms^-2).

Answer 1

u = 10 ms^-1; θ = 30°; T = 5s; g = 10 ms^-2

∴ v_x = u cos 30° = 10 x \(\frac{\sqrt{3}}{2}\) = 5\(\sqrt{3}\) ms^-1

v_y = u sin 30° = 10 x \(\frac{1}{2}\) = 5 ms^-1

= -25 + 125

= 100 m