Question

A rock thrown with speed 7.50 m/s and launch angle 30.0 ∘ (above the horizontal) travels a horizontal distance of d = 18.0 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:

height from where rock was thrown is 27.916 m

Explanation:

speed = 7.50 m/s

θ = 30°

g= 9.8 m/s²

horizontal distance = 18 m

time require for vertical displacement

$time = \frac{distance}{velocity} \\t = \frac{18}{7.5\ cos30^0}$

t = 2.8 sec

now for calculation of height

s = ut + 0.5 a t²

-h = v sinθ× t + 0.5 ×(-9.8)× (2.8²)

-h = 7.5 sin30°× 2.8 - 0.5 ×(9.8)× (2.8²)

-h = -27.916 m

h= 27.916 m

height from where rock was thrown is 27.916 m