Question

A projectile is fired at time t= 0.0 s from point 0 at the edge of a cliff, with initial velocity components of Vox = 30 m/s and Voy = 100 m/s. The projectile
rises, and then falls into the sea at point P. The time of flight of the projectile is 25 s. Assume air resistance is negligible.
15.
0
+
What is the magnitude of the velocity at time t = 15.0 s?
O 56 m's​

Answer 1

Answer:

At t = 15.0 s the magnitude of the velocity is 58.31 m/s

Explanation:

The given parameters are;

V₀ₓ = 30 m/s

$V_{0y}$ = 100 m/s

The time of flight of the projectile = 25 s

For projectile motion;

Vₓ = V₀ × cos(θ₀)

The magnitude of the velocity V = √(V₀ₓ² + $V_{0y}$²)

We have the magnitude of the initial velocity = √(30² + 100²) = 10·√109 m/s

cos(θ₀) = V₀ₓ/V₀ = 30/(10·√109) = 3/√109

θ₀ = cos⁻¹(3/√109) = 73.3°

The components of the velocity after time t is given by the relations;

Vₓ = V₀ × cos(θ₀) = 30 m/s

$V_y$ =  V₀ × sin(θ₀) - g×t

When $V_y$ = 0, we have;

0 =  V₀ × sin(θ₀) - g×t

g×t  =  V₀ × sin(θ₀)  = 10·√109×0.958 = 100 m/s

t = 100/g = 100/10 = 10 s

The time to reach maximum height = 10 s

At 15.0 seconds, we have;

$V_y$ =  V₀ × sin(θ₀) - g×t = 10·√109×0.958  - 10×15 = -50 m/s

Therefore, the projectile is returning at 50 m/s

The magnitude of the velocity =√(30² + 50²) = 10·√34 m/s = 58.31 m/s.