Question

Projectile A is launched horizontally at a speed of 20. Meters per second from the top of a cliff and strikes a level surface below, 3.0 seconds later. Projectile B is launched horizontally from the same location at a speed of 30. Meters per second. Approximately how high is the cliff?

Answer 1

consider the motion of projectile A in vertical direction :

v₀ = initial velocity of projectile A in vertical direction = 0 m/s         (since the projectile was launched horizontally)

a = acceleration of the projectile = g = acceleration due to gravity = 9.8 m/s²

t = time of travel for projectile A = 3.0 seconds

Y = vertical displacement of projectile A = height of the cliff = h = ?

using the kinematics equation along the vertical direction as

Y = v₀ t + (0.5) a t²

h = (0) (3.0) + (0.5) (9.8) (3.0)²

h = 44.1 m

Answer 2

Answer:

The height of the cliff is $44.1\;\rm{m$.

Explanation:

Given: Projectile $A$ is launched horizontally at a speed of $20$ Meters per second from the top of a cliff and strikes a level surface below, $3.0$ seconds later. Projectile $B$ is launched horizontally from the same location at a speed of $30$ meter per second.

Using the formula of motion: $S=ut+\frac{1}{2}at^2$

Substiuting all the values: $u=0\;\rm{ms^{-1}, a=9.8\;\rm{ms^{-2}\;\& \;t=3.0\; seconds$

Let $h$ be the height of the cliff.

$h=0\times3+0.5\times(9.8)\times3^2\\h=44.1\;\rm{m$

Hence, height of the cliff is $44.1\;\rm{m$.

Learn more about Laws of motion here:

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