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?

Answer 1

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.