If a 1450kg vehicle drives horizontally off a 45m high cliff at 85km/h, at what speed and at what angle would the car hit the ground. Express both of your final answers to two decimal places.
So far I have gotten time as 3.03s. Distance on the x axis as 71.54m and vertical velocity as -29.69m/s. I got my final angle to be -0.89 by using TanTheta=verticalvel/horizontalvel but I don't know if this is correct.
To solve this problem, we can use the principles of kinematics, specifically the equations of motion.
First, we need to find the initial velocity of the vehicle when it drives off the cliff. Since the vehicle is horizontally driven, there is no vertical component contributing to the initial velocity. Therefore, the initial velocity in the x-direction is the same as the velocity in the x-direction throughout the motion. Given that the speed of the vehicle is 85 km/h, we need to convert it to m/s:
85 km/h = 85 * 1000 m/3600 s = 23.61 m/s
Now, let's find the time it takes for the vehicle to hit the ground. We can use the equation: h = (1/2) * g * t^2, where h is the initial height (45 m), g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. Rearranging the equation, we have:
t = sqrt(2h/g)
t = sqrt(2 * 45 m / 9.8 m/s^2) ≈ 3.03 s (rounded to two decimal places)
Next, we can find the horizontal distance traveled by the vehicle during this time. We know that the initial velocity in the x-direction is 23.61 m/s, and the time is 3.03 s. Using the equation: d = v * t, where d is the distance and v is the velocity, we have:
d = 23.61 m/s * 3.03 s ≈ 71.54 m (rounded to two decimal places)
Now, let's find the vertical velocity just before the vehicle hits the ground. We can use the equation: v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time. The initial vertical velocity is 0 m/s since there is no vertical velocity in the beginning, and the acceleration due to gravity is -9.8 m/s^2 (negative due to downwards direction). We have:
v = 0 m/s + (-9.8 m/s^2) * 3.03 s ≈ -29.69 m/s (rounded to two decimal places)
Now, let's find the angle at which the vehicle hits the ground. We can use the formula: tan(θ) = vertical velocity / horizontal velocity. Plugging in the values:
tan(θ) = -29.69 m/s / 23.61 m/s ≈ -1.26
To find the angle θ, we can use the arctan function (inverse tangent):
θ ≈ arctan(-1.26) ≈ -48.50 degrees (rounded to two decimal places)
Since angles are conventionally measured in the counterclockwise direction from the positive x-axis, a negative angle suggests a direction opposite to that convention. Therefore, the angle at which the vehicle hits the ground is approximately 48.50 degrees below the negative x-axis.
To summarize the final answers:
- The vehicle hits the ground with a speed of approximately 23.61 m/s (rounded to two decimal places).
- The vehicle hits the ground at an angle of approximately -48.50 degrees (rounded to two decimal places).