A stunt driver wants to make his car jump over 8 cars parked side by side below

Question

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

A With what minimum speed must he drive off the horizontal ramp? The vertical height of the ramp is 1.5 mabove the cars and the horizontal distance he must clear is 22 m.

Express your answer using two significant figures.

Part B

If the ramp is now tilted upward, so that "takeoff angle" is 9.0 above the horizontal, what is the new minimum speed?(Figure 1)

Express your answer using two significant figures.

Explanation / Answer

1)

Find the time needed for an object to fall 1.5 m

s = ut + at^2 / 2

where

s = distance = -1.5 m (negative because final height below initial height)

u = initial vertical velocity = 0 m/s

t = time = ?

a = acceleration by gravity = -9.8 m/s^2

-1.5 = 0t + (-9.8)t^2/2

t = 0.5533 s

Now find the horizontal velocity.

r = vt

where

r = horizontal range = 22 m

v = horizontal velocity = ?

t = time = 0.5533 s

22 = v x 0.5533

v = 39.76 m/s

2)

Horizontal component is:

Vh = Vcos(9)

Vertical component is:

Vv = Vsin(9)

Horizontal range is:

r = vt

where

r = horizontal range = 22

v = horizontal velocity = Vh = Vcos(9)

t = time = ?

r = Vcos(9) x t

V = 22 / (cos(9) x t)

Vertical displacement is:

s = ut + at^2/2

where

s = distance = -1.5 m (negative because final height below initial height)

u = initial vertical velocity = Vv = Vsin(9)

t = time = ?

a = acceleration by gravity = -9.8 m/s^2

-1.5 = Vsin(9)t + (-9.8)t^2/2

4.9t^2 - sin(9)tV - 1.5 = 0

4.9t^2 - sin(9)t22 / (cos(9)t)) - 1.5 = 0

4.9t^2 - 22tan(9) - 1.5 = 0

t = 1.009 s

V = 22 / (cos(9) × 1.009)

V = 22.08 m/s

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