Question

Learning Goal: To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 12.0 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 15.5 m before hitting the ground. From what height was the rock thrown? Use the value g = 9.800 m/s2 for the free-fall acceleration.

Answer 1

Answer:

The height from which the rock was thrown is 1.92 m

Solution:

As per the question:

Speed with which the rock is thrown, v = 12.0 m/s

Horizontal distance traveled by the rock before it hits the ground, d = 15.5 m

Launch angle, $\theta = 30.0^{\circ}$

Now,

To calculate the height, h from which the rock was thrown:

First, since we consider the horizontal motion in the trajectory of the rock, thus the time taken is given by:

$t = \frac{d}{vcos\theta}$

$t = \frac{15.5}{12.0cos30.0^{\circ}} = 1.49\ s$

Now,

The height from which the rock was thrown is given by the kinematic eqn, acceleration in the horizontal direction is zero:

$h = vsin\theta t - \frac{1}{2}gt^{2}$

$h = 12.0sin30.0^{\circ}\times 1.49 - \frac{1}{2}\times 9.8\times 1.49^{2} = - 1.92\ m$

Answer 2

Answer:

the height was thrown = 1.938 m

Explanation:

given,

speed of the rock = 12 m/s

angle of launch = 30.0 ∘

horizontal distance = d = 15.5 m

acceleration due to gravity  = 9.8 m/s²

$u_x = u cos \theta$

$u_x = 12 cos 30^0$

$u_x = 10.39 m/s$

$u_y = u sin \theta$

$u_y = 12 sin 30^0$

$u_y = 6 m/s$

$time = \dfrac{s_x}{u_x}$

$time = \dfrac{15.5}{10.39}$

t = 1.49 s

$s_y = u_y t + \dfrac{1}{2}gt^2$

$s_y = 6\times 1.49 - \dfrac{1}{2}\times 9.8 \times 1.49^2$

$s_y = -1.938 m$

hence, the height was thrown = 1.938 m