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.

Step-by-step 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+(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+(1)/(2)*9.8* t^2 \\ \implies t=\sqrt{(10.0*2)/(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*1.43 \\ =7.1\text{ m} \end{gathered}`

Final answer:

Thus the correct answer is option B.