An arrow is shot from a height of 1.5 m toward a cliff of height H . It is shot with a velocity of 30 m/s at an angle of 60º above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?

Solution 34PE

We are required to calculate the height of the cliff, the maximum height reached by the arrow and the speed of the arrow before it reaches the cliff.

(a)

Step 1 of 5</p>

The angle projection

Velocity of projection m/s

The height from which the arrow is shot m

The height of the cliff

Time required to travel to the top edge of the cliff s

The x-component of the projection velocity

The y-component of the projection velocity

Step 2 of 5</p>

Since the arrow is shot from a height of 1.5 m, hence from the kinematic equation (negative acceleration since the arrow is shot upward),

m

m

Therefore, the height of the cliff is 27.02 m.

(b)

Step 3 of 5</p>

The equation for maximum height attained by a projectile is,

m

This maximum height will be attained by the projectile when thrown from 1.5 m above the ground.

Therefore, the height of the projectile from the ground would be m

The height of the projectile from the ground is m

(c)