Question

A marble is thrown horizontally with a speed of 15 m/sec from the top of a building. When it strikes the ground, the marble has a velocity that makes an angle of 65⁰ with the horizontal. From what height above the ground was the marble thrown?

Answer 1

Answer:

52.73978 m

Explanation:

$v_x$ = 15 m/s

$\theta$ = 65°

The x component will remain the same. So, we can use the relation

$tan\theta=\dfrac{v_y}{v_x}\\\Rightarrow v_y=v_xtan\theta\\\Rightarrow v_y=15\times tan65\\\Rightarrow v_y=32.1676\ m/s$

From the following equation

$v_y^2=u_y^2+2as\\\Rightarrow s=\dfrac{v_y^2-u_y^2}{2a}\\\Rightarrow s=\dfrac{32.1676^2-0^2}{2\times 9.81}\\\Rightarrow s=52.73978\ m$

The height above the ground is 52.73978 m