In the figure below, a stone is projected at a cliff of height h with an initial speed of 45.0 m/s directed 60.0° above the horizontal. The stone strikes at A, 5.20 s after launching.
Find the speed of the stone just before impact at A.
Find the maximum height H reached above the ground.
To find the speed of the stone just before impact at point A, we can use the equations of projectile motion. Let's break this down into steps:
Step 1: Compute the vertical and horizontal components of the initial velocity.
Given:
Initial speed (v₀) = 45.0 m/s
Launch angle (θ) = 60.0°
To find the vertical component (vᵥ) of the initial velocity, we use trigonometry:
vᵥ = v₀ * sin(θ)
vᵥ = 45.0 * sin(60.0°)
vᵥ ≈ 38.93 m/s
To find the horizontal component (vₕ) of the initial velocity, we use trigonometry as well:
vₕ = v₀ * cos(θ)
vₕ = 45.0 * cos(60.0°)
vₕ ≈ 22.50 m/s
Step 2: Determine the time taken to reach the highest point (t).
Since the stone strikes at point A after 5.20 s, and we assume symmetry in the projectile's trajectory, it would reach the highest point halfway through the total time of flight:
t = 5.20 s / 2
t = 2.60 s
Step 3: Find the vertical component of velocity at the highest point (vᵥmax).
At the highest point of the projectile's trajectory, the vertical component of velocity becomes zero since the object momentarily stops before reversing direction. Therefore, we have:
vᵥmax = 0 m/s
Step 4: Calculate the maximum height reached (H).
Using the equation of motion, we can find the maximum height above the ground:
vᵥmax = vᵥ - g * t
0 = 38.93 - 9.81 * t
9.81 * t = 38.93
t = 38.93 / 9.81
t ≈ 3.97 s
To find the maximum height (H), we substitute the value of t into the equation of motion:
H = vᵥmax * t - 0.5 * g * t²
H = 0 * 3.97 - 0.5 * 9.81 * (3.97)²
H ≈ 76.86 m
Therefore, the speed of the stone just before impact at point A is approximately 38.93 m/s, and the maximum height reached above the ground is approximately 76.86 m.