GIVEN: We launched a toy car off a ramp (the launcher has 4 different positions)

Question

GIVEN:

We launched a toy car off a ramp (the launcher has 4 different positions) and recorded the distance traveled. We need to find the initial and final velocities of the car at each position.

Ramp height: 31 inches

Ramp length: 6.2 inches

Angle : 19.94 degrees

Launch position (1,2,3,4) :

1 = 26.5 inches traveled

2 = 35.1 inches traveled

3 = 45.6 inches traveled

4 = 57.7 inches traveled

SOLVE FOR INITIAL LAUNCH VELOCITY AT THE LAUNCH POINT (V0)

SOLVE FOR LANDING VELOCITY OFF THE RAMP (Vf)

please clearly show all equations and variables. please solve using only the given variables

Explanation / Answer

given, toy car is launched off the ramp

ramp height = ho = 31 inches

ramp length, l = 6.2 in

angle, theta = 19.94 deg

so initial height of launch, h = ho + lsin(theta) = 33.114 in

now, let the horizontal distance travelled be x

then initial velocity be v

so time of flight = t

h = vsin(theta)*t - 0.5gt^2

and

x = vcos(theta)*t

so, -h = vsin(theta)*x/vcos(thtea) - 0.5g*x^2/v^2cos^2(theta)

gx^2/2*cos^2(theta)(tan(theta)*x + h) = v^2

plugging in values

v^2 = 18.2189x^2/(0.362x + 33.114)

also final landing velocity Vf

so from conservation of energy

0.5v^2 + gh = 0.5vf^2

vf^2 = v^2 + 2gh = v^2 + 2132.5416

so for the 4 settings given

1. x = 26.5 in

vi = 17.308 in/s

vf = 49.3165 in/s

2. x = 35.1 in

vi = 22.132 in/s

vf = 51.209 in/s

3. x = 45.6 in

vi = 28.586 in/s

vf = 54.3115 in/s

4. x = 57.7 in

vi = 33.514 in/s

vf = 57.059 in/s

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