<u>Answer:</u>
a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s
b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s
<u>Explanation:</u>
a) The height of ramp = 1.5 meter
Horizontal distance he must clear = 22 meter
The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.
We have equation of motion,
, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.
In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8
, we need to calculate time when displacement = 1.5 meter.
![1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds](https://tex.z-dn.net/?f=1.5%3D0%2At%2B%5Cfrac%7B1%7D%7B2%7D%20%2A9.8%2At%5E2%5C%5C%20%5C%5C%20t%20%3D%200.553%20seconds)
So the car has to cover a distance of 22 meter in 2.119 seconds.
So minimum speed required = 22/0.553 = 39.78 m/s
Minimum speed must he drive off the horizontal ramp = 39.78 m/s
b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V
So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
Substituting
![-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0](https://tex.z-dn.net/?f=-1.5%3D0.122V%2At-%5Cfrac%7B1%7D%7B2%7D%20%2A9.8%2At%5E2%5C%5C%20%5C%5C%204.9t%5E2-0.122Vt-1.5%3D0)
In case of horizontal motion
Horizontal speed of car = V cos 7 = 0.993V
So it has to travel 22 meter in t seconds
0.993Vt = 22, Vt = 22.155 m
Substituting in the equation ![4.9t^2-0.122Vt-1.5=0](https://tex.z-dn.net/?f=4.9t%5E2-0.122Vt-1.5%3D0)
We will get ![4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds](https://tex.z-dn.net/?f=4.9t%5E2-0.122%2A22.155-1.5%3D0%5C%5C%20%5C%5C%20t%20%3D%200.926%20seconds)
Speed required = 22.155/0.926 = 23.93 m/s
Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s