Kyle, a stunt driver, wants to drive a motorcycle off a 4.0m high parkade and into a lake. He needs a horizontal range of 13m in order to land in water deep enough to keep him safe. With what speed must he leave the parkade?
We can solve this problem using the principles of projectile motion.
Let's consider the vertical and horizontal motions separately:
Vertical motion:
We know that the initial vertical velocity is 0 m/s (since Kyle is starting from the top of the parkade) and the vertical displacement is -4.0m (negative because Kyle is falling downwards). We also know that the acceleration due to gravity is approximately 9.81 m/s^2. Using the following kinematic equation:
vf^2 = vi^2 + 2*a*d
where:
vf = final vertical velocity (0 m/s when Kyle hits the water)
vi = initial vertical velocity (unknown)
a = acceleration due to gravity (-9.81 m/s^2)
d = vertical displacement (-4.0m)
By rearranging the equation and plugging in the known values, we can solve for the initial vertical velocity:
0 = vi^2 + 2*(-9.81)*(-4.0)
0 = vi^2 + 78.48
vi^2 = 78.48
vi ≈ 8.86 m/s
Horizontal motion:
We know that the horizontal range (R) is 13m and the horizontal velocity is the unknown we are trying to find. We can use the following kinematic equation for horizontal motion:
R = vx*t
where:
vx = horizontal velocity (unknown)
t = time of flight (unknown)
To find the time of flight, we can use the vertical motion formula:
vf = vi + a*t
0 = 8.86 + (-9.81)*t
-8.86 = -9.81*t
t ≈ 0.90s
Now we can use the time of flight to solve for the horizontal velocity:
13 = vx*0.90
vx ≈ 14.44 m/s
Therefore, Kyle must leave the parkade with a speed of approximately 14.44 m/s in order to reach the lake safely.