Question

A projectile is shot from the edge of a cliff 125 m above ground level with an initial speed of 65.0 m/s at an angle of 37 degrees with the horizontal. (a) determine the time taken by the projectile to hit the point P at ground level. (b) determine the range X of the projectile as measured from the base of the cliff at the instant just before the projectile hits point P. Find (x) the horizontal and vertical components of its velocity and (d) the magnitude of the velocity and (e) the angle made by the velocity vector with the horizontal (f) Find the maximum height above the cliff top reached by the projectile

Answer 1

Answer:

(a) The time it takes the projectile to hit the ground is approximately 10.42 seconds

(b) The range of the projectile from the base of the cliff is approximately 540.92 meters

(c) The horizontal component of the velocity is approximately 51.91 m/s

The vertical component of the velocity is approximately 39.12 m/s

(d) The magnitude of the velocity of the projectile is 65.0 m/s

(e) The angle made by the velocity vector of the projectile with the horizontal is 37°

(f) The maximum height above the cliff top reached by the projectile is approximately 77.99 meters

Explanation:

The height of the cliff above ground level, y₀ = 125 m

The initial speed of the projectile, u = 65.0 m/s

The angle of elevation of the projectile, θ = 37°

(a) The time for the projectile to hit the ground, t, is given as follows;

y = y₀ + u·sin(θ)·t -(1/2)·g·t²

g = 9.81 m/s²

At ground level, y = 0, we get;

0 = 125 + 65·sin(37)·t - 4.905·t²

t = (-65·sin(37) ± √((65·sin(37))² - 4×(-4.905)×125))/(2 × (-4.905))

∴ t ≈ -2.45 or t ≈ 10.42

The time it takes the projectile to hit the ground, t ≈ 10.42 seconds

(b) The range of the projectile X = u·cos(θ) × t

∴ X = 65.0 × cos(37°) × 10.42 ≈ 540.92

The range of the projectile from the base of the cliff, X ≈ 540.92 meters

(c) The horizontal component of the velocity, uₓ = u×cos(θ)

∴ uₓ = 65.0 ×cos(37°) ≈ 51.91

The horizontal component of the velocity, uₓ ≈ 51.91 m/s

The vertical component of the velocity, $u_y$ = u×sin(θ)

∴ $u_y$ = 65.0 ×sin(37°) ≈ 39.12

The vertical component of the velocity, $u_y$ ≈ 39.12 m/s

(d) The magnitude of the velocity, $\left | u \right |$, is the given speed  = 65.0 m/s

$\left | u \right |$ = √($u_y$² + uₓ²)

∴ $\left | u \right |$ = √(39.12² + 51.91²) = 65

$\left | u \right |$ = 65.0 m/s

(e) The angle made by the velocity vector with the horizontal = The angle of elevation with which the projectile is launched = 37°

(f) The maximum height above the cliff top reached by the projectile, $y_{max \ of \ cliff}$, is given as follows;

$y_{max \ of \ cliff}= \dfrac{ \left(u \times sin(\theta)\right)^2}{2 \cdot g}$

∴ $y_{max \ of \ cliff}$ = (65.0 × sin(37°))²/(2 × 9.81) ≈ 77.99

$y_{max \ of \ cliff}$ ≈ 77.99 meters.