A human cannonball is shot from a cannon at a speed of 21 meters per second at an angle of 20 degrees; how long before his height is 0? How far did he travel in that time?
To determine how long it takes for the human cannonball's height to reach 0, we need to consider the vertical motion of the cannonball. We can analyze the motion using the formula for vertical displacement:
y = y0 + v0y * t - 0.5 * g * t^2
Where:
y = final vertical displacement (0 meters in this case)
y0 = initial vertical displacement (unknown)
v0y = initial vertical velocity (v0 * sin θ)
t = time taken
g = acceleration due to gravity (approximately 9.8 m/s^2)
Since we know the initial vertical velocity is v0 * sin θ and the initial vertical displacement is 0 (since the cannonball starts at the ground), we can simplify the formula:
0 = 0 + (v0 * sin θ) * t - 0.5 * g * t^2
Now, we can solve this quadratic equation for t. Rearranging the equation gives:
0.5 * g * t^2 - (v0 * sin θ) * t = 0
Dividing both sides of the equation by t gives:
0.5 * g * t - v0 * sin θ = 0
Applying the quadratic formula, we have:
t = (v0 * sin θ) / (0.5 * g)
Now, let's substitute the given values into the formula:
v0 = 21 m/s
θ = 20 degrees
g = 9.8 m/s^2
t = (21 * sin 20) / (0.5 * 9.8)
Using a calculator, we can find t approximately equal to 1.34 seconds.
Next, to determine the horizontal distance traveled (x), we can use the formula for horizontal displacement:
x = v0x * t
Where:
x = horizontal displacement (unknown)
v0x = initial horizontal velocity (v0 * cos θ)
t = time taken (1.34 seconds)
Since we know the initial horizontal velocity is v0 * cos θ, we can substitute the values into the formula:
v0 = 21 m/s
θ = 20 degrees
x = (21 * cos 20) * 1.34
Using a calculator, we find x approximately equal to 19.25 meters.
Therefore, the human cannonball will reach the ground after approximately 1.34 seconds, and in that time, they will travel approximately 19.25 meters horizontally.