Given data
*The given mass of the car is m = 885 kg
*The given speed is u = 40 m/s
*The given acceleration of the car is a = 3 m/s^2
*The given distance is s = 200 m
*The height of the cliff is H = 120 m
(a)
The formula for the time taken by the car is given by the equation of motion as
[tex]s=ut+\frac{1}{2}at^2[/tex]
Substitute the values in the above expression as
[tex]\begin{gathered} 200=(40)t+\frac{1}{2}\times3\times t^2 \\ 200=40t+1.5t^2 \\ t=4.30\text{ s} \end{gathered}[/tex]
Hence, the time taken by the car is t = 4.30 s
(b)
The formula for the time taken by the car to land is given as
[tex]T=\sqrt[]{\frac{2H}{g}}[/tex]
*Here g is the acceleration due to the gravity
Substitute the values in the above expression as
[tex]\begin{gathered} T=\sqrt[]{\frac{2\times120}{9.8}} \\ =4.94\text{ s} \end{gathered}[/tex]
Hence, the time taken by the car to land is T = 4.94 s
(c)
The formula for the horizontal distance from the cliff is given as
[tex]R=u\times T[/tex]
Substitute the values in the above expression as
[tex]\begin{gathered} R=(40)(4.94) \\ =197.6\text{ m} \end{gathered}[/tex]
Hence, the horizontal distance from the cliff is R = 197.6 m
