Answer to Question #296448 in Mechanics | Relativity for yan

Question

A stone is thrown upward from the top of a building at an angle of 30.0° to the

horizontal and with an initial speed of 20.0 m/s. The point of release is 45.0 m above the

ground.

a. How long does it take to hit the ground?

b. Find the stone’s speed impact.

c. Find the horizontal range of the stone.

Answer 1

Explanations & Calculations

• This is a comprehensive questionnaire to understand the projectile motion under gravity.

a)

• Apply $s=ut +\frac{1}{2}at^2$ upwards on the stone for its motion starting at the throw until the impact. So at the impact, it would have travelled a displacement of $\small -45.0\,m$. And the equation to solve would be like

$\qquad\qquad \begin{aligned} \small \uparrow\qquad -45&=\small (20.0\sin30)t+\frac{1}{2}(-9.8)t^2 \end{aligned}$

• You can solve the quadratic equation formed here and find the time of flight. Neglect any negative values....[4.21s]

b)

• Apply energy conservation with respect to the impact level.

$\qquad\qquad \begin{aligned} \small mgh+\frac{1}{2}mv_1^2&=\small mgh+\frac{1}{2}mv_2^2\\ \small mg(45)+\frac{1}{2}m(20)^2&=\small 0+\frac{1}{2}.m.v_2^2\\ \small 9.8\times45+\frac{1}{2}.(20)^2&=\small \frac{v_2^2}{2} \end{aligned}$

• Solve for $\small v_2$ .

c)

• The all-time constant horizontal velocity is $\small v\cos\theta=20\cos30$
• Then the range can simply be calculated by

$\qquad\qquad \begin{aligned} \small \to \qquad s&=\small ut\\ &=\small 20\cos30\times \text{time of flight} \end{aligned}$