Ebo throws a ball in to the air.it's velocity at the start is 18m/s at an angle 37° to the ground. Calculate
A,the range and height of the ball
B,the time taken when the ball lands
AAAaannndd the bot gets it wrong yet again!
For a body with initial velocity v at an angle θ,
Range = v^2/g sin2θ = 18^2/9.8 sin74° = 31.78 m
Height = v^2/(2g) sin^2θ = 18^2/19.6 sin^2(37°) = 5.987 m
time to land = 2v/g sinθ = 36/9.8 sin37° = 2.21 s
I apologize for the mistake in the previous answer. You are correct that the range, height, and time taken for the ball to land are calculated differently.
Using the correct equations:
Range = v^2/g sin2θ = 18^2/9.8 sin74° = 31.78 m
Height = v^2/(2g) sin^2θ = 18^2/19.6 sin^2(37°) = 5.987 m
Time taken to land = 2v/g sinθ = 2*18/9.8 sin37° = 2.21 s
Thank you for bringing this to my attention and I hope this answer is satisfactory.
To calculate the range and height of the ball, we need to separate the initial velocity (18 m/s) into its vertical and horizontal components.
First, let's find the vertical component of the initial velocity:
Vertical component = Velocity * sin(angle) = 18 * sin(37°) ≈ 10.84 m/s
Next, let's find the time it takes for the ball to reach its peak height:
The vertical motion of the ball is subject to gravity, so we can use the equation:
vf = vi + at
where vf is the final vertical velocity (which becomes 0 at the peak height), vi is the initial vertical velocity, a is the acceleration due to gravity (-9.8 m/s²), and t is the time.
0 = 10.84 - 9.8t
9.8t = 10.84
t ≈ 1.11 seconds
Now, let's calculate the peak height (H):
Using the equation:
H = vi * t + (1/2) * a * t²
H = 10.84 * 1.11 + (1/2) * -9.8 * (1.11)²
H ≈ 6.04 meters
To calculate the range (R):
We can use the equation:
Range = Horizontal component * Time of flight
The horizontal component is given by:
Horizontal component = Velocity * cos(angle) = 18 * cos(37°) ≈ 14.42 m/s
Time of flight = 2 * t (the ball takes the same amount of time to reach the peak and fall back down)
Time of flight = 2 * 1.11 ≈ 2.22 seconds
Now we can calculate the range:
Range = 14.42 * 2.22 ≈ 32.03 meters
So, the answers to the questions are:
A) The range of the ball is approximately 32.03 meters, and the height of the ball is approximately 6.04 meters.
B) The time taken when the ball lands is approximately 2.22 seconds.
A) To calculate the range and height of the ball, we need to use the equations of motion. Since the initial velocity has a horizontal and vertical component, we can separate them as follows:
Horizontal motion:
- Acceleration is zero
- Distance (range) = velocity x time
- Therefore, the range of the ball is R = V0x * t, where V0x is the horizontal component of the initial velocity, and t is the time it takes for the ball to land.
Vertical motion:
- Acceleration is -9.81 m/s² (acceleration due to gravity)
- Final velocity (v) is zero when the ball reaches maximum height and also when it lands
- We can use the following equation to find the maximum height (h) of the ball: h = (v0y)² / (2 * g), where v0y is the vertical component of the initial velocity.
First, let's find the horizontal and vertical components of the initial velocity:
V0x = V0 * cos(theta) = 18 * cos(37) = 14.32 m/s
V0y = V0 * sin(theta) = 18 * sin(37) = 10.80 m/s
Now, let's find the maximum height:
v = 0, v0 = V0y = 10.80 m/s, a = -9.81 m/s²
h = (v0²) / (2 * a) = (10.80²) / (2 * 9.81) = 6.12 m (rounded to two decimal places)
Finally, let's find the range:
R = V0x * t. To find t, we can use the equation y = v0y * t + (1/2)*a*t². At the instant the ball reaches its maximum height, y = h. Since the ball will reach the same height on its way down, we can plug in y = h and solve for t.
h = v0y * t + (1/2)*a*t²
6.12 = 10.80 * t + (1/2)*(-9.81)*t²
0 = 4.905t² + 10.80t - 6.12
Using the quadratic formula, we get:
t = 1.29 s (rounded to two decimal places)
Therefore, the range of the ball is R = V0x * t = 14.32 * 1.29 = 18.48 m (rounded to two decimal places).
B) The time taken when the ball lands is 2t, since it takes the same amount of time to reach maximum height and to fall back down to the ground. Therefore, the time taken when the ball lands is 2 * 1.29 = 2.58 s (rounded to two decimal places).