Answer to Question #243873 in Physics for Ulysses

Question

6. Jackstanding on top of a building 15 m above the ground, drops sandbag. At the same instant, Jill aims and throws a rock towards the sandbag at a velocity of . Jill is positioned 3.5 m away from Jack and her arm as she throws the rock is positioned 1.5 m above the ground. () At what angle should fill throw the rock in order to hit the sandbag? (b) At what time will the rock hit the sandbag? (c) At whut height with respect to the ground does the collision (rock and sandbag) occur (d) What is the velocity of the rock when it hits the sandbag?

Answer 1

The height where the rock will hit the sandbag:

$$h=\frac{v_{yf}^2-v_{yi}^2}{2g}=(H-y)-d=(15-1.5)-d.$$

The time the impact will occur:

$$t=\frac{v_{yf}-v_{yi}}{g},\\\space\\ v_{if}=v_{yi}-gt.$$

During this time, the bag will be falling and will cover a distance before the impact:

$$d=\frac{gt^2}{2}=13.5-\frac{v_{yf}^2-v_{yi}^2}{2g}.$$

During this time, the rock will need to cover the horizontal distance of 3.5 m:

$$v_xt=R=3.5\text{ m},\\\space\\ v_x=\frac{v_{yi}}{\tan\theta},\\\space\\ \frac{v_{yi}t}{\tan\theta}=R.$$