A STUNTMAN DRIVES A MOTORCYCLE OFF A 350 m CLIFF GOING 70 mph. THE ANGLE OF ELEVATION OF THE CLIFF IS 21 HE IS HOPING TO MAKE IT ACROSS A 261 M WIDE RIVER AND LAND ON A LEDGE 82 M HIGH. DOES HE MAKE IT?

Answer 1

First, let's convert the given information into meters and seconds to maintain consistency of units.

1 mph = 0.44704 m/s, so 70 mph = 31.293 m/s

Now, we can break down the initial velocity into horizontal and vertical components:

Initial horizontal velocity (Vx) = initial velocity * cos(angle) = 31.293 m/s * cos(21 degrees) = 29.363 m/s
Initial vertical velocity (Vy) = initial velocity * sin(angle) = 31.293 m/s * sin(21 degrees) = 11.225 m/s

We must now find how long it takes for the stuntman to reach the ledge's height (82 m). To do this, we can use the following equation:

final vertical position = initial vertical position + (initial vertical velocity * time) - (0.5 * g * time^2)
g = 9.81 m/s^2 (acceleration due to gravity)

82 = 0 + (11.225 * time) - (0.5 * 9.81 * time^2)

Now, solve the quadratic equation for time (t):

9.81t^2 - 11.225t + 82 = 0

This gives us two solutions, however one of them will be negative and invalid for this situation.

t ≈ 3.143 s

Now that we know the time it takes to reach the ledge's height, let's find out how far horizontally the stuntman will travel in that time:

horizontal distance = initial horizontal velocity * time = 29.363 m/s * 3.143 s ≈ 92.25 m

As the distance required to cross the river is 261 m, the stuntman does not make it to the other side.