A ball is thrown from a building standing 30m high, landing 10m away from the building 3.2 seconds after it was thrown. Find the angle above the horizontal and the initial velocity of the ball
0 = 30 + Vi (3.2) - 4.9 (3.2)^2
Vi = 6.31 up
10 = u (3.2)
u = 3.13
tan theta = 6.31/3.13
theta = 63.6 degrees up from horizontal
speed = sqrt(3.13^2 + 6.31^2)
= 7.04
To find the angle above the horizontal and the initial velocity of the ball, we can break down the problem into two components: the horizontal (x) component and the vertical (y) component.
Let's start with the horizontal component:
1. Horizontal Component:
- The ball lands 10m away from the building, so we know the horizontal displacement (Δx) is 10m.
- We also know the time taken (t) is 3.2 seconds.
We can use the equation: Δx = v₀ * cos(θ) * t,
where v₀ is the initial velocity of the ball and θ is the angle above the horizontal.
Plugging in the values we have:
10m = v₀ * cos(θ) * 3.2s.
Now, let's find the vertical component:
2. Vertical Component:
- The building is 30m high, which gives us the initial height (h₀) of the ball.
- We know the time taken (t) is 3.2 seconds.
- The acceleration due to gravity (g) is approximately 9.8 m/s².
We can use the equation: h = h₀ + v₀ * sin(θ) * t - (1/2) * g * t²,
where h is the final vertical displacement of the ball.
Plugging in the values we have:
0m = 30m + v₀ * sin(θ) * 3.2s - (1/2) * 9.8 m/s² * (3.2s)².
Solving the equations simultaneously will give us the values of θ and v₀.
Remember, the velocity of the ball has two components: horizontal (v₀ * cos(θ)) and vertical (v₀ * sin(θ)). The magnitude of the initial velocity can be calculated using the Pythagorean theorem:
v₀ = √[(v₀ * cos(θ))² + (v₀ * sin(θ))²].