Baseball launch from edge of cliff at initial velocity of 40m/s at an angle 20 degrees below the horizontal. The cliff height of 200m above ground, baseball lands on elevated platform 20m above ground

Question

a) Draw a diagram showing the initial height, final height, initial position, final position, initial velocity (with component), and the baseball's trajectory.

b) Find the time it takes for the baseball to land, in seconds.

c) Find the horizontal distance traveled by the baseball, in meters.

d) Find the baseball's velocity just before hitting the platform, in m/s.

Answer 1

I choose to make the coordinate plane be located at the initial position. This makes the bottom of the cliff (final height) be - 200m.

The time it takes can be calculated by considering the y-direction of travel which starts with an initial velocity of 40sin(20°) downward so -13.68 m/s and accelerates at -9.8 m/s²:

x_f = x_i + v_it + 1/2at²

-200 = 0 + -13.68t + 1/2(-9.8)t²

-4.9t² - 13.68t + 200 = 0

Solve using the quadratic equation:

t = 5.1435 and t = -7.9355

We can ignore the negative time answer so t = 5.1435 seconds or (rounded) t = 5 seconds

The horizontal distance is calculated using the horizontal velocity multiplied by the time:

x_f = x_i + v_it

x_f = 0 + 40cos(20°)(5.1435) = 193 meters or rounded x_f = 200 m

The final velocity has an x-component of 40cos(20°) = 37.59 m/s and it has a y-component calculated as:

v_f = v_i + at

v_f = -13.68 + (-9.8)(5.1435) = -64.09 m/s

With an x-component of 37.59 m/s and a y-component of -64.09 m/s we can find the vector velocity. The magnitude of the velocity is √(37.59² + (-64.09)²) = 74.3 m/s (I'll let you round if you want) and a direction calculated using tan^-1(64.09/37.59) = 59.6° below the horizontal