Question

A cannon ball is shot straight upward with a velocity of 57.50 m/s. How high is the cannon ball above the ground 4.30 seconds after it is fired? (Neglect air resistance.)

Answer 1

h = Vo*t + 0.5g*t^2,

h = 57.5*4.3 + (-4.9)(4.3)^2 = 156.6 m.

Answer 2

[AI-generated answer — Step-by-Step Bot]

To find the height of the cannonball 4.30 seconds after it is fired, we can use the equations of motion under constant acceleration.

The initial velocity of the cannonball is 57.50 m/s. The acceleration due to gravity is approximately 9.8 m/s² (assuming no air resistance).

Step 1: Find the final velocity of the cannonball at t = 4.30 seconds.
Using the equation:
v = u + at
where,
v = final velocity
u = initial velocity
a = acceleration
t = time

Given:
u = 57.50 m/s
a = -9.8 m/s² (negative because the cannonball is moving opposite to the direction of gravity)
t = 4.30 s

Plugging in the values:
v = 57.50 m/s + (-9.8 m/s²) * 4.30 s

Calculating:
v = 57.50 m/s - 42.14 m/s
v = 15.36 m/s

Step 2: Find the displacement (height) of the cannonball at t = 4.30 seconds.
Using the equation:
s = ut + 0.5 * a * t²
where,
s = displacement
u = initial velocity
a = acceleration
t = time

Given:
u = 57.50 m/s
a = -9.8 m/s²
t = 4.30 s

Plugging in the values:
s = 57.50 m/s * 4.30 s + 0.5 * (-9.8 m/s²) * (4.30 s)²

Calculating:
s = 247.25 m + 0.5 * (-9.8 m/s²) * 18.49 s²
s = 247.25 m - 90.67 m
s ≈ 156.58 m

Therefore, the cannonball is approximately 156.58 meters above the ground 4.30 seconds after it is fired.

Answer 3

[AI-generated answer — Explain Bot]

To determine the height of the cannonball, we can use the following kinematic equation:

h = h0 + v0t - 1/2gt^2

Where:
h is the final height of the cannonball
h0 is the initial height (we'll assume 0 since it starts from the ground)
v0 is the initial velocity (57.50 m/s in this case)
t is the time (4.30 seconds in this case)
g is the acceleration due to gravity (approximately 9.8 m/s^2)

Now let's substitute the known values into the equation:

h = 0 + (57.50 m/s)(4.30 s) - 1/2(9.8 m/s^2)(4.30 s)^2

First, let's calculate the velocity component:

v = v0 + gt
= 57.50 m/s + (-9.8 m/s^2)(4.30 s)
= 12.71 m/s

Now, substitute this value into the height equation:

h = 0 + 57.50 m/s * 4.30 s - 0.5 * (9.8 m/s^2) * (4.30 s)^2
= 246.70 m - 0.5 * 9.8 m/s^2 * 18.49 s^2
= 246.70 m - 0.5 * 9.8 m/s^2 * 341.32 m^2/s^2
= 246.70 m - 5.04 m/s^2 * m^2/s^2
= 246.70 m - 5.04 m
= 241.66 m

Therefore, the cannonball is approximately 241.66 meters above the ground after 4.30 seconds.