Question

A man stands on top of a 25 m tall cliff. The man throws a rock upwards at a 25 degree angle above the horizontal with an initial velocity of 7.2 m/s. What will be the vertical position of the rock at t = 2 seconds? What will be the horizontal position of the rock at that time? (Show all work)

Answer 1

Answer:

The vertical position of the rock is 38.96 m and the horizontal position of the rock is 13 m

Explanation:

Velocity of the rock = 7.2 m/s

Horizontal component of the velocity

$v_x=7.2\cos25^\circ$

$v_x=7.2\times 0.9$ m/s

$v_x=6.5$ m/s

Vertical component of the velocity

$v_y=7.2\sin 25^\circ$

$v_y=7.2\times 0.42$ m/s

$v_y=3.02$ m/s in the upwards direction

After t = 2 seconds, the vertical displacement

Using second equation of motion

$h=ut+\frac{1}{2}gt^2$

$Y=-3.02\times 2+\frac{1}{2}\times 10\times 2^2$

$\implies Y=20-6.04$

$\implies Y=13.96$ m

Since the man is already on the top of 25 m cliff

Therefore total vertical position

$=25+13.96$

$=38.96$ m

The horizontal displacement in 2 seconds

$X=6.5\times 2$

$X=13$m

Thus the vertical position of the rock is 38.96 m and the horizontal position of the rock is 13 m

Hope this helps.