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yaroslaw [1]
3 years ago
6

Learning Goal: To practice Problem-Solving Strategy 4.1 for projectile motion problems. A rock thrown with speed 12.0 m/s and la

unch 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.
Physics
2 answers:
lesantik [10]3 years ago
8 0

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

DiKsa [7]3 years ago
6 0

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

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