Question

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

Answer 1

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

Answer 2

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).

Answer 3

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.