Answer:
The maximum range [tex]R_{max}= 132. 72[/tex] m
Explanation:
Given,
The initial velocity of the car, u = 30 m/s
The height of the cliff, h = 50 m
Let the car drives off the cliff with a horizontal velocity of 30 m/s.
The formula for a projectile that is projected from a height h from the ground is given by the relation
[tex]R_{max}= \frac{u}{g}\sqrt{u^{2} + 2gh }[/tex] m
Where,
g - acceleration due to gravity
Substituting the values in the above equation
[tex]R_{max}= \frac{30}{9.8}\sqrt{30^{2} + 2X9.8X50 }[/tex]
= 132.72 m
Hence, the car lands at a distance, [tex]R_{max}= 132. 72[/tex] m