You are the stunt director for a testosterone-laden action movie. A car drives up a ramp inclined at 10.0° above the horizontal, reaching a speed of 40.0 m/s at the end. It will jump a canyon that is 101 meters wide. The lip of the takeoff ramp is 201 meters above the floor of the canyon. (a) How long will the car take to cross the canyon? (b) What is the maximum height of the cliff on the other side so that the car lands safely? (c) What angle with the horizontal will the car's velocity make when it lands on the other side? Assume the height of the other side is the maximum value you just calculated. Express the angle as a number between −180° and +180°.
a)2.56s
b)186.7m
To answer these questions, we will break them down step by step.
(a) How long will the car take to cross the canyon?
To calculate the time it takes for the car to cross the canyon, we can use the horizontal component of its velocity. The horizontal component remains constant throughout the motion.
Given:
- Ramp angle (θ) = 10.0°
- Width of the canyon (d) = 101 meters
- Speed of the car (v) = 40.0 m/s
We can start by calculating the horizontal component of the car's velocity (v_x):
v_x = v * cos(θ)
Substituting the given values:
v_x = 40.0 m/s * cos(10.0°)
Using a calculator, we find that v_x is approximately 39.74 m/s.
To find the time (t) taken to cross the canyon, we can use the formula:
t = d / v_x
Substituting the given values:
t = 101 meters / 39.74 m/s
Calculating this, we find that the car will take approximately 2.54 seconds to cross the canyon.
(b) What is the maximum height of the cliff on the other side so that the car lands safely?
To find the maximum height of the cliff, we need to consider the vertical motion of the car. The car will follow a projectile path, reaching its maximum height before descending back to the ground.
Given:
- Height of the takeoff ramp (h) = 201 meters
- Acceleration due to gravity (g) = 9.8 m/s^2
The time it takes for the car to reach its maximum height can be calculated using the vertical component of its initial velocity (v_y0) and the acceleration due to gravity.
v_y0 = v * sin(θ)
Substituting the given values:
v_y0 = 40.0 m/s * sin(10.0°)
Using a calculator, we find that v_y0 is approximately 6.95 m/s.
Using the equation for vertical displacement, we can find the time it takes for the car to reach its maximum height:
h = v_y0 * t - (1/2) * g * t^2
Simplifying the equation for t:
(1/2) * g * t^2 - v_y0 * t + h = 0
This is a quadratic equation with respect to t. Solving this equation, we find two solutions for t. The positive solution corresponds to the time taken for the car to reach its maximum height:
t = (-b + √(b^2 - 4ac)) / (2a)
Substituting the values into the quadratic equation:
t = (-(-6.95) + √((-6.95)^2 - 4 * (0.5 * 9.8) * 201)) / (2 * (0.5 * 9.8))
Using a calculator, we find that the positive solution is approximately 1.6 seconds. This is the time taken for the car to reach its maximum height.
To find the maximum height (H) of the cliff, we can use the vertical component of the car's velocity (v_y):
v_y = v_y0 - g * t
Substituting the values:
v_y = 6.95 m/s - 9.8 m/s^2 * 1.6 s
Calculating this, we find that v_y is approximately -6.04 m/s. The negative sign indicates that the car is moving downwards at this point.
Using the equation for vertical displacement, we find:
H = v_y * t + (1/2) * g * t^2
Substituting the values:
H = -6.04 m/s * 1.6 s + (1/2) * 9.8 m/s^2 * (1.6 s)^2
Calculating this, we find that the maximum height of the cliff on the other side is approximately 3.68 meters.
(c) What angle with the horizontal will the car's velocity make when it lands on the other side? Assume the height of the other side is the maximum value you just calculated. Express the angle as a number between −180° and +180°.
To find the angle at which the car's velocity makes with the horizontal when it lands on the other side, we can use trigonometry.
Using the known values:
- Vertical velocity (v_y) = -6.04 m/s (downwards)
- Horizontal velocity (v_x) = v_x (calculated earlier)
- Angle with the horizontal (θ')
We can calculate θ' using the equation:
θ' = arctan(v_y / v_x)
Substituting the values:
θ' = arctan(-6.04 m/s / 39.74 m/s)
Using a calculator, we find that θ' is approximately - 8.68 degrees.
Since the car is moving downwards, the angle is negative. Therefore, the car's velocity makes an angle of approximately -8.68 degrees with the horizontal when it lands on the other side.
Note: The angle is between -180° and +180°, so it falls within this range.