A stunt driver wants to make his car jump over eight 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 1.5m above the cars, and the horizontal distance he must clear is 20m . wich is 36m/s but...
b)If the ramp is now tilted upward, so that "takeoff angle" is 14° above the horizontal, what is the new minimum speed???? how do I do this?
how long to fall 1.5 meters?
1.5 = .5 (9.81) t^2
how far horizontal ? 20 meters
so
u t = 20 where t = t from the vertical problem above
u = 20/that t
now do your vertical problem with vi = s sin 14 and u = s cos 14
h = vi t - 4.9 t^2 = 1.5
1.5 = (s sin 14)t - 4.9 t^2
20 = (s cos 14) t
so t = 20/s cos 14
1.5 = sin 14*20/ cos 14 -4.9(400/s^2 cos^2 14)
solve for s
how do i solve for s? I don't undestand!!!
1.5 = 4.99 - 2000/s^2
-3.49 = -2000/s^2
s^2 = 573
s = 24 m/s
To solve this problem, we can use the principles of projectile motion and apply them to the scenario given.
a) For the first part of the question, where the ramp is horizontal, we need to find the minimum speed at which the car must be driven off the ramp to clear the 20m distance and reach a height of 1.5m above the cars parked below.
To solve for the minimum speed, we can use the following kinematic equation:
y = y0 + v0y * t - (1/2) * g * t^2
where:
y = vertical displacement (1.5m)
y0 = initial vertical position (0m)
v0y = initial vertical velocity
g = acceleration due to gravity (-9.8 m/s^2)
t = time of flight
Since y0 = 0, the equation simplifies to:
y = v0y * t - (1/2) * g * t^2
We can also use the formula for horizontal displacement:
x = v0x * t
where:
x = horizontal displacement (20m)
v0x = initial horizontal velocity
Since v0x is constant and there is no acceleration horizontally, we can solve for v0x:
v0x = x / t
Now, we can eliminate time by rearranging the equation for horizontal displacement:
t = x / v0x
Substituting this expression for time into the equation for vertical displacement, we get:
y = v0y * (x / v0x) - (1/2) * g * (x / v0x)^2
Plugging in the values for y (1.5m), x (20m), and g (-9.8 m/s^2), we can solve for v0y.
1.5 = v0y * (20 / v0x) - (1/2) * (-9.8) * (20 / v0x)^2
Simplifying this equation, we get:
0 = 9.8 * (20 / v0x)^2 - 30 / v0x
To find v0x, we can make use of the quadratic formula:
v0x = (-b ± √(b^2 - 4ac)) / (2a)
where:
a = 9.8
b = -30
c = 0
Plugging in these values, we solve the quadratic equation for v0x. The positive root will give us the minimum speed needed to clear the distance.
v0x = 36.94 m/s (rounded to two decimal places)
Therefore, the minimum speed the stunt driver must drive off the horizontal ramp is 36.94 m/s.
b) Now, let's consider the case where the ramp is tilted upward at an angle of 14° above the horizontal.
To calculate the new minimum speed, we need to break down the initial velocity into its horizontal and vertical components. The vertical component will need to provide enough upward velocity to clear the 1.5m obstacle while the horizontal component will need to provide enough velocity to cover the 20m distance:
v0y = v0 * sin(θ)
v0x = v0 * cos(θ)
where:
v0y = initial vertical velocity
v0x = initial horizontal velocity
v0 = initial velocity magnitude
θ = takeoff angle (14°)
Since the vertical motion is affected by gravity, we can use the same equation as in part a):
1.5 = v0y * t - (1/2) * g * t^2
Using the equation for vertical velocity and substituting it in the equation above, we obtain:
1.5 = v0 * sin(θ) * t - (1/2) * g * t^2
The horizontal displacement equation remains the same:
x = v0 * cos(θ) * t
Now, we need to solve the equations simultaneously to eliminate the time variable. Rearranging the equation for time:
t = x / (v0 * cos(θ))
Substituting this expression into the equation for vertical displacement:
1.5 = (v0 * sin(θ) * x) / (v0 * cos(θ)) - (1/2) * g * (x^2 / (v0 * cos(θ))^2)
Simplifying the equation:
1.5 = x * tan(θ) - (1/2) * g * (x^2 / (v0 * cos(θ))^2)
Rearranging this equation to isolate v0:
v0 = sqrt((g * x^2) / (2 * cos(θ) * (tan(θ) * x - 2 * y)))
Plugging in the values of g = 9.8 m/s^2, x = 20m, and y = 1.5m, we can solve for v0:
v0 = 38.44 m/s (rounded to two decimal places)
Therefore, the new minimum speed required for the ramp tilted upward at a 14° takeoff angle is 38.44 m/s.