Question

An arrow is shot from a height of 1.3 m toward a cliff of height H . It is shot with a velocity of 34 m/s at an angle of 58.1º above the horizontal. It lands on the top edge of the cliff 4.1 s later. What is the height of the cliff?

Answer 1

The height of the cliff is 39.655 m

Given:

Height at which the arrow was shot, h = 1.3 m

Velocity of the arrow, u = 34 m/s

Angle,
`\theta' = 58.1^(\circ)`

Time of the fight, t = 4.1 s

Solution:

Let the Height of the cliff be H

Since, the motion of the object is projectile motion and the direction of motion is vertical at some angle

Therefore, we consider the vertical component of velocity,
`u_(y) = usin\theta`.

Now,

The Height of the cliff is given by applying the second equation of motion in the projectile:

Thus

`s = u_(y)t - (1)/(2)gt^(2)`

`s = 34sin60^(\circ)* 4.1 - (1)/(2)* 9.8* 4.1^(2)`

s = 38.355 m

Now, the height of the cliff, H:

H = s + h = 38.355 + 1.3 = 39.655 m