A human cannonball is launched from a cannon at 26.4 m/s at 20.4 degrees above the horizontal. (Assume he/she lands in a net the same height as the cannon.) How high in the air does he/she go?
viy=26.4*sin20.4
at the top, vyfinal=0
vyfinal=viy-g*t
solve for time t, then h=viy-1/2 g t^2
To determine how high the human cannonball goes, we need to analyze the vertical motion of the projectile. We can use the following kinematic equation:
y = y0 + v0y * t - (1/2) * g * t^2
Where:
- y is the vertical position/height of the human cannonball
- y0 is the initial vertical position (which is zero since we are measuring the height above the ground)
- v0y is the initial vertical velocity
- t is the time
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
First, let's find the initial vertical velocity (v0y). We can use trigonometry to split the initial velocity (26.4 m/s) into its horizontal and vertical components.
v0y = v0 * sin(theta)
Where:
- v0 is the initial velocity (26.4 m/s)
- theta is the launch angle (20.4 degrees)
Now, we can solve for the time it takes for the human cannonball to reach its maximum height. At the maximum height, the vertical velocity is zero.
0 = v0y - g * t
Rearranging the equation, we can solve for t:
t = v0y / g
Finally, we can substitute the value of t we found into the equation for the vertical position:
y = y0 + v0y * t - (1/2) * g * t^2
Since the human cannonball returns to the same height as the cannon (y0 = 0), the equation simplifies to:
y = v0y * t - (1/2) * g * t^2
Substituting the values of v0y, g, and t:
y = (v0 * sin(theta)) * (v0 * sin(theta) / g) - (1/2) * g * (v0 * sin(theta) / g)^2
Now, we can plug in the known values and calculate the height (y):
v0 = 26.4 m/s
theta = 20.4 degrees
g = 9.8 m/s^2
Finally, calculate the value of y to find the height.