A man stands on the roof of a building of height 15.8m and throws a rock with a velocity of magnitude 25.2m/s at an angle of 32.7∘ above the horizontal. You can ignore air resistance.
Part A
Calculate the maximum height above the roof reached by the rock.
y =
9.46mCorrect
Part B
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
h = 15.8 + (25.2 sin 32.7°)t - 4.9t^2
= 15.8 + 13.61t - 4.9t^2
I get max height = 25.25 at t=1.388, which is as you say, 9.46m above the roof.
h=0 when t=3.66
v = 13.61 - 9.8t = -22.26 when h=0
To solve this problem, we can use the equations of motion for projectile motion. Let's break down the problem step by step:
Part A
To calculate the maximum height above the roof reached by the rock, we need to find the vertical displacement of the rock. We can use the equation:
y = y0 + v₀y * t + 0.5 * a * t²
where:
y = vertical displacement (we don't know yet)
y₀ = initial vertical position (height of the building, 15.8m)
v₀y = initial vertical velocity (v₀ * sin(θ), where v₀ is the magnitude of the velocity, 25.2m/s, and θ is the angle, 32.7 degrees)
a = acceleration due to gravity (-9.8m/s²)
t = time taken to reach the maximum height (we don't know yet)
Since the rock reaches its maximum height when it stops moving vertically (its vertical velocity becomes zero), we can find the time taken to reach the maximum height using the vertical motion equation:
v = v₀y + a * t
where:
v = vertical velocity at the maximum height (0m/s)
Using this equation, we can rearrange it to solve for t:
t = (v - v₀y) / a
Plugging in the given values, we have:
t = (0 - (25.2 * sin(32.7))) / -9.8
Simplifying the equation, we get:
t ≈ 1.556s
Now, we can substitute the value of t into the equation for y to find the vertical displacement:
y = 15.8 + (25.2 * sin(32.7)) * 1.556 + 0.5 * -9.8 * (1.556)²
Simplifying the equation, we get:
y ≈ 9.46m
Therefore, the maximum height above the roof reached by the rock is approximately 9.46m.
Part B
To calculate the magnitude of the velocity of the rock just before it strikes the ground, we can use the horizontal motion equation:
x = x₀ + v₀x * t
where:
x = horizontal displacement (we don't know yet)
x₀ = initial horizontal position (0m, since the rock is thrown from the roof)
v₀x = initial horizontal velocity (v₀ * cos(θ), where v₀ is the magnitude of the velocity, 25.2m/s, and θ is the angle, 32.7 degrees)
t = time taken to reach the maximum height (1.556s, as calculated in Part A)
Since the horizontal displacement is the same at both the starting point and the landing point (since they are at the same height), we can set x to 0. Solving for v₀x, we have:
v₀x = -x₀ / t
Plugging in the given values, we have:
v₀x = 0 / 1.556
The value of v₀x is 0 as the horizontal displacement is 0 when the rock returns to the ground.
Thus, the magnitude of the velocity of the rock just before it strikes the ground is 0 m/s.