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Question

# A rock is thrown from the top of a $20-\mathrm{m}$ building at an angle of $53^{\circ}$ above the horizontal. If the horizontal range of the throw is equal to the height of the building, with what speed was the rock thrown? What is the velocity of the rock just before it strikes the ground?

Solution

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In this problem, a rock is thrown at an angle of $\theta = 53^\circ$ on top of a $h=20 \text{ m}$ building as shown in the figure below.

To solve this problem, we can use the formulas:

$x = x_0 + v_0 t + \frac{1}{2}at^2 \tag{1}$

Also, the acceleration in a projectile motion are as follows:

$a_y = -g\text{ and } a_x = 0 \tag{2}$

Also, the horizontal and vertical components of the initial velocity can be solved as follows:

$v_{0,x}= v_0 \cos \theta \tag{3}$

$v_{0,y}= v_0 \sin \theta \tag{4}$ ## Recommended textbook solutions #### Fundamentals of Physics

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