basketball shot at height 1.8m (above ground) with initial velocity 10m/s and release angle 52degrees (horizontal. if basket rim is 3.05m above ground, will ball have sufficient vertical height and go through the basket which is 10m away from shot taken? much appreciated
To determine if the basketball will clear the rim and go through the basket, we need to find the maximum height the ball will reach and whether it will still be above the rim at the point where it is 10m away from the shot taken.
First, let's break down the problem into horizontal and vertical components.
Horizontal component:
The horizontal distance traveled by the basketball can be determined using the formula:
distance = velocity * time
In this case, the initial velocity (v₀) of the basketball is 10m/s, and the time it takes to reach the basket (10m away) can be calculated using the equation:
distance = velocity * time
10m = 10m/s * time
time = 1 second
Therefore, it will take 1 second for the basketball to reach the basket.
Vertical component:
We can use the following equations of motion to analyze the vertical component:
1. Vertical velocity at any point (vₓ) can be found using:
vₓ = v₀ * sin(angle)
where v₀ is the initial velocity (10m/s) and the angle is the release angle (52°).
2. The vertical displacement (dₓ) can be determined using:
dₓ = v₀ * sin(angle) * time - (1/2) * g * time²
where g is the acceleration due to gravity (approximately 9.8m/s²). It should be noted that we use the negative sign in the second term of the equation since the acceleration due to gravity opposes the upward motion of the ball.
3. The maximum height (h) reached by the ball can be calculated by finding the vertical displacement (dₓ) when the vertical velocity (vₓ) becomes zero. This occurs at the peak of the trajectory.
0 = v₀ * sin(angle) - g * time
v₀ * sin(angle) = g * time
h = dₓ = (v₀ * sin(angle) * time) - (1/2) * g * time²
Let's compute the values:
vₓ = 10m/s * sin(52°) ≈ 7.68m/s
dₓ = (10m/s * sin(52°) * 1s) - (1/2) * 9.8m/s² * (1s)² ≈ 4.11m
h = dₓ = (10m/s * sin(52°) * 1s) - (1/2) * 9.8m/s² * (1s)² ≈ 4.11m
Therefore, the ball will reach a maximum height of approximately 4.11m above the ground.
Now, let's see if the ball clears the rim. The height of the rim is 3.05m above the ground, so the ball will need to have a vertical height greater than 3.05m to clear it.
Since the maximum height is approximately 4.11m, the ball will easily clear the rim.
In conclusion, given the initial conditions, the basketball will have sufficient vertical height to clear the rim and go through the basket, which is 10m away from the shot taken.