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
![s=ut+(1)/(2)at^2](https://img.qammunity.org/2023/formulas/physics/college/obvwlaq2brhvduordk5kuedw6274qqa7x1.png)
Substitute the values in the above expression as
![\begin{gathered} 200=(40)t+(1)/(2)*3* t^2 \\ 200=40t+1.5t^2 \\ t=4.30\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/8ebfm237prvduv18hlzfykyrnj77xxyxxv.png)
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
![T=\sqrt[]{(2H)/(g)}](https://img.qammunity.org/2023/formulas/physics/college/symzfmk6t069tfvvhrniyuwtykgxxmaq6w.png)
*Here g is the acceleration due to the gravity
Substitute the values in the above expression as
![\begin{gathered} T=\sqrt[]{(2*120)/(9.8)} \\ =4.94\text{ s} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/uv9q62b0vl49xmuebtrao9wxk14i63dd38.png)
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
![R=u* T](https://img.qammunity.org/2023/formulas/physics/college/q6i2yh50l8hju282tf1cynd15m9ecwacgw.png)
Substitute the values in the above expression as
![\begin{gathered} R=(40)(4.94) \\ =197.6\text{ m} \end{gathered}](https://img.qammunity.org/2023/formulas/physics/college/nu105pswrxs79ig9tb2bsuiv3vynxtupa8.png)
Hence, the horizontal distance from the cliff is R = 197.6 m