A boy stands on the edge of a building 7 meters high and throws a rock at 60-degree angle. The release point from his hand is 1.6 meters above his feet(so the release point above the ground is 8.6 meters.) The rock lands on the ground 15 meters away from the building at a point level with the base of the building. Assuming no air resistance, what was the initial velocity of the throw? (assume gravity is 9.81m/s^2)
I'm stuck on how to solve. I think I need to figure out how long the rock is in the air, which would be time it takes for the rock to arrive at the apex of the throw and then drop from the apex height to the ground. But since I don't know the initial velocity, how do I determine the height of the apex?
The horizontal distance away that it landed after T seconds tells you that
Vo cos 60 T = 15 , which leads to
T = 30/Vo
Also, the vertical velocity component Vo sin 60 is related to the total time T in the air by
Vo sin 60 T - (g/2) T^2 = -8.6
You have two equations in two unknowns.
Substitue 30/Vo for T in the last equation and solve for Vo
thanks, I get it.
To solve this problem and find the initial velocity of the throw, you can use the kinematic equations of motion. First, let's find the height of the apex.
Since the boy throws the rock at a 60-degree angle, the vertical component of the initial velocity is given by V₀y = V₀ * sin(θ), where θ is the launch angle (60 degrees) and V₀ is the initial velocity.
To determine the height of the apex, we can apply the equations of motion to the vertical motion of the rock:
1. h = 8.6 m (release point above the ground)
2. V₀y = V₀ * sin(θ)
3. g = 9.81 m/s² (acceleration due to gravity)
4. Time of flight = t (time it takes for the rock to reach the ground)
Now we will use the equation for vertical displacement:
h = V₀y * t - (1/2) * g * t²
By substituting the given values, we can solve for t:
8.6 = V₀ * sin(60) * t - (1/2) * 9.81 * t²
Now, solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b² - 4ac)) / (2a)
where a = -4.905 (1/2 * 9.81), b = V₀ * sin(60) and c = -8.6.
By substituting the values of a, b, and c into the quadratic formula, you will get two values for t. Since time cannot be negative, you should choose the positive value as the time it takes for the rock to reach the ground.
Once you have the value of t, you can find the horizontal distance traveled by the rock using the equation:
Range = V₀ * cos(θ) * t,
where cos(θ) is the horizontal component of the initial velocity.
Substituting the values for θ, t, and the given range (15 meters), you can solve for V₀:
15 = V₀ * cos(60) * t.
By rearranging the equation, you can solve for V₀:
V₀ = 15 / (t * cos(60)).
By plugging in the value of t from earlier, you will find the initial velocity of the throw.