A jeep starts from rest with a constant acceleration of 4m/s2.At the same time a car travels with a constant speed of 36km/h overtake and passes the jeep how far beyond the starting point will the jeep overtakes the car?

25m

Explanation:

Let's assume the Jeep attains a velocity of 36km/h ; a constant speed same with that of the car.

While the Jeep is accelerating to that speed, the car with that speed passes it.

Now we can calculate the time taken for the Jeep to attain the velocity of 36km/h on her constant acceleration.

This time is t = v/a; from Newton's Law of Motion:

a = V-U / t ; a-acceleration

V is final velocity = 36km/h

U is initial velocity 0 since the body starts from rest.

Hence t = 36000/3600 ÷ 4 = 2.5s

Note conversting from km/h to m/s we multiply by 1000/3600.

But the distance covered by the car while the Jeep just accelerates is

S = U × t = 10× 2.5 = 25m.

Note From Newton's law of Motion, distance for constant speed is defined as: U × t

Hence the Car would be 25m off the starting point just as the Jeep accelerates. It would overtake the Jeep when it just covers 25m from the Jeep starting point.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

the answer would be a