Question

A projectile is launched horizontally from a cliff that is 10.0 m above with the ground an initial velocity of 5.00m/s. How far from the base of the cliff does it land

Answer 1

Given:

The initial horizontal velocity of the projectile, u_x=5.00 m/s

The initial vertical velocity of the projectile, u_y=0 m/s

The height of the cliff, h=10.0 m

To find:

The range of the fight of the projectile.

Explanation:

From the equation of motion, the height of the cliff through which the projectile falls during its flight is given by,

$h=u_yt+\frac{1}{2}gt^2$

Where g is the acceleration due to gravity and t is the time of flight.

On substituting the known values in the above equation,

$\begin{gathered} 10.0=0+\frac{1}{2}\times9.8\times t^2 \\ \implies t=\sqrt{\frac{10.0\times2}{9.8}} \\ =1.43\text{ s} \end{gathered}$

The range of the projectile, that is, the distance between the point where the projectile lands and the base of the cliff is given by,

$x=u_xt$

On substituting the known values,

$\begin{gathered} x=5.00\times1.43 \\ =7.1\text{ m} \end{gathered}$

Final answer:

Thus the correct answer is option B.