A rock is thrown at an angle of $ 42^\circ $ above the horizontal at a speed of $ 14 \, \text{m/s} $. Determine how long it takes the rock to hit the ground. • Nerd Notes

AP Physics

Unit 1 - Vectors and Kinematics

Intermediate

Mathematical

MCQ

A rock is thrown at an angle of $42^\circ$ above the horizontal at a speed of $14 \, \text{m/s}$ . Determine how long it takes the rock to hit the ground.

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Step	Derivation/Formula	Reasoning
1	$v_{i_y} = v_i \sin(\theta)$	Calculate the initial vertical component of the velocity. Use the sine function since the angle given is above the horizontal.
2	$v_{i_y} = 14 \sin(42^\circ)$	Substitute the given values into the equation.
3	$v_{i_y} \approx 14 \cdot 0.6691$	Calculate the sine of 42 degrees to get approximately 0.6691.
4	$v_{i_y} \approx 9.3674 \text{ m/s}$	Calculate the initial vertical velocity component.
5	$\Delta y = v_{i_y} t + \frac{1}{2} a t^2$	Use the kinematic equation for vertical displacement. Here, $\Delta y$ is zero because the rock returns to the same vertical level.
6	$0 = 9.3674 t + \frac{1}{2} (-9.81) t^2$	Since the acceleration due to gravity $a$ is -9.81 m/s $^2$ , substitute the known values into the equation.
7	$0 = 9.3674 t – 4.905 t^2$	Simplify the equation by multiplying out the terms.
8	$0 = t (9.3674 – 4.905 t)$	Factor out the common term $t$ .
9	$t = 0$ or $t = \frac{9.3674}{4.905}$	This gives two solutions, $t = 0$ (initial time) and $t$ when the rock hits the ground.
10	$t \approx \frac{9.3674}{4.905}$	Solve for $t$ by dividing the terms.
11	$t \approx 1.91 \text{ seconds}$	Calculate the final value of $t$ .
12	$\boxed{1.91 \text{ seconds}}$	This answer closely matches choice (b)

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AP Physics

Unit 1 - Vectors and Kinematics

Intermediate

Mathematical

MCQ

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