A man stands on the roof of a 19.0 m-tall building and throws a rock with a velocity of magnitude 30.0 m/sat an angle of 42.0 degrees above the horizontal. You can ignore air resistance.
Calculate the maximum height above the roof reached by the rock.
Calculate the magnitude of the velocity of the rock just before it strikes the ground.
Calculate the horizontal distance from the base of the building to the point where the rock strikes the ground.
The vertical component of velocity when the rock is thrown is 30 sin 42 = 20.07 m/s
Calculate the time T that it takes the rock to hit the ground using that initial velocity.
Hint: 20.07 T - (g/2) T^2 + 19.0 = 0
I assume you know what g is, and how to solve that quadratic equation. Take the positive root.
T times the horizontal velocity component is the horizontal distance that the rock travels.
The maximum height h of the rock is attained when the vertical velocity component is zero. This happens when
gt = 20.07 m/s
h = (1/2)(20.07 m/s)^2/g
Ebergy considerations can be used to compute the velocity at impact
Vo = 30 m/s
Xo = 0
Yo = 0
theta = 42
ax = 0 m/s^2 i
ay = -9.8 m/s^2 j
Vox = 30cos42
Vox = ____ m/s i
To find the maximum height reached by the rock above the roof, we need to analyze the projectile motion of the rock. We'll break down the motion into horizontal and vertical components.
First, let's find the time it takes for the rock to reach its highest point. We can use the vertical component of the initial velocity. The initial vertical velocity (Vy) can be calculated using the magnitude of the velocity (30.0 m/s) and the angle (42.0 degrees) above the horizontal.
Vy = (30.0 m/s) * sin(42.0 degrees)
Next, we can use the equation of motion to find the time it takes to reach the highest point. At the highest point, the vertical velocity becomes zero.
0 = Vy - g * t_max_height
Where g is the acceleration due to gravity (9.8 m/s^2) and t_max_height is the time to reach the highest point.
Solving for t_max_height:
t_max_height = Vy / g
Now we can find the maximum height (h_max) above the roof by using the equation of motion for vertical displacement:
h_max = Vy * t_max_height - (1/2) * g * t_max_height^2
To calculate the magnitude of the velocity just before the rock strikes the ground, we'll use the equation of motion for vertical velocity:
Vfy = Vy - g * t_total
Where t_total is the total time of flight for the projectile. We can determine t_total by finding the time it takes for the rock to hit the ground. Assuming the ground is at the same height as the base of the building, we can use the equation:
h_max = (1/2) * g * t_total^2
Solving for t_total:
t_total = sqrt((2 * h_max) / g)
Finally, we can calculate Vfy:
Vfy = Vy - g * t_total
Lastly, to find the horizontal distance from the base of the building to the point where the rock strikes the ground, we'll use the equation of motion for horizontal displacement:
x_max = Vx * t_total
Where Vx is the horizontal component of the initial velocity. We can calculate Vx using the magnitude of the velocity (30.0 m/s) and the angle (42.0 degrees) above the horizontal:
Vx = (30.0 m/s) * cos(42.0 degrees)
Now we can calculate x_max.