Stunt rider question

Question

A stunt rider is propelled upward from his motorbike by a spring loaded ejector seat. The rider was travelling horizontally at 60km h^-1 when the ejector seat was triggered, and as they leave the seat they are travelling with a vertical velocity if 15m s^-1. The seat is 1.0m off the ground.

a) What is the initial velocity of the stunt rider (in km h^-1)?

b) How high does the stunt rider reach?

c) How far along the track does the stunt rider land on the ground?

d) What is the velocity of the stunt rider when they hit the ground (in km h^-1)

Answer 1

Before starting to solve the problem, determine what is known from the problem, what is wanted, and define the coordinate system. I usually recommend drawing a picture since it makes it easier to visualize. In this problem, the positive x-direction will be in the direction the motorbike is traveling and the positive y-direction is up.

Initial values:

Horizontal: x₀ = 0; v_0x = 60 km/hr; a_x = 0

Vertical: y₀ = 1 .0 m; v_0y = 15 m/s; a_y = g = -9.8 m/s²

All calculations assume that air resistance can be neglected.

(a) Initial velocity of stunt rider:

The first step in solving this problem is to convert the vertical velocity 15 m/s into km/hr. This is a commonly used conversion factor, so you may want to write it down. However, it is also important to know how to find the conversion factor since you won't always have it on a test.

The conversion from m/s to km/hr (or the reverse) rests on knowing two other relations: 1000 m = 1 km and 3600 s = 1 hour. These can each be written as a fraction equal to 1.

so (15 m/s) (1 km/1000 m)(3600 s/1 hr) = (combining numerators and denominators)

[(15)(1)(3600)(m ^. km ^. s)] / [(1000) (s ^. m ^. hr)] = 54 km/hr (NOTE: this is a lot easier to see if you write each term as a vertical fraction, then cancel common factors and complete the multiplication and division)

Now that the vertical velocity is in km/hr. The next step is to use the Pythagorean Theorem to find the velocity.

V = sqrt[ (v_0x)² + (v_0y)² ] = sqrt[ (60 km/hr)² + (54 km/hr)² ] = 80.7 km/hr or 81 km/hr (to 2 significant figures)

(b) How high does he go?

Any question about height requires only the equation of motion in the vertical. Since we don't have the time of flight, we can use another equation:

v² = v_o² + 2a(y - y_o). (This equation comes from combining the position and velocity equations to eliminate time.) Solving this equation for y (position) and using the fact that at the top of the arc (maximum height) the vertical velocity, v, is zero, and the acceleration, a = g get:

y = y₀ - (v_o² ) / (2 g) = 1.0 m + [(15 m/s)² / (2*9.8 m/s²)] = 12.5 m (13 m to 2 sig. figs.)

(c) How far along the track?

This part of the question happens only in the x-direction, except the time in the air is determined by the motion in the y-direction. Finding the time is easy if the object starts and stops at the same level - solve for the time to max height and multiply by 2. However in this case, he starts above the ground and ends at the ground, so have to use the distance equation of motion in the y-direction to solve for the time. y = y₀ + v_oy t + (1/2) a t².

This is a quadratic equation in t, so can solve by putting it into the standard form then using the quadratic formula. The revised equation becomes:

(1/2) g t² - v_oy t - y₀ = 0 (or - (1/2) g t² + v_oy t + y₀ = 0 )

Use the coefficients of the t terms in the quadratic formula (use the absolute value of g since the negative considered in the signs of the revised equation).

t = [ v_oy +/- sqrt[ (v_oy)² - (4)(1/2)(g)(y₀)] / [2 (1/2) g]

= {15 =/- sqrt[ 15² + 4(9.8)]} / 9.8 = [(15 +/- 16.2) / 9.8] s (use the positive root since elapsed time cannot be negative) = 3.2 s

Use this time in the distance equation for x: x = x₀ + v_ox t + (1/2) a_x t² = v_ox t

since time is in seconds, need to convert v_ox from km/hr to m/s, use the same method/conversions as in part (a) except use inverses, because going in the opposite direction. 60 km/hr = 16.7 m/s

x = (16.7 m/s) ( 3.2 s) = 53 m down the track.

(d) Speed when lands (in km/hr):

The velocity component in the x-direction is unchanged: 60 km/hr

The velocity component in the y-direction is found using the time of travel: v = v₀ + a t = v₀ - g t = (15m/s) - (9.8 m/s²)(3.2 s) = -16.4 m/s = 59 km/hr

Use the Pythagorean Theorem to solve for the total speed: sqrt[(60 km/hr)² + (59 km/hr)² ] = 84 km/hr