A body is projected vertically upward with a speed 40 m/s. The distance travelled by the body in the last second of the upward journey is [take g = 9.8 m/s² and neglect effect of air resistance]

(1) 4.9 m

(2) 9.8 m

(3) 12.4 m

(4) 19.6 m

A body is projected vertically upward with a speed 10 m/s and another at the same time with the same speed in the downward direction from the top of a tower. The magnitude of acceleration of first body with respect to second is {take g = 10 m/s²}

(1) Zero

(2) 10 m/s²

(3) 5 m/s²

(4) 20 m/s²

A car travelling at a speed of \(30\) km/h is brought to rest at a distance of \(8\) m by applying brakes. If the same car is moving at a speed of \(60\) km/h, then it can be brought to rest with the same brakes in:
1. \(64\) m
2. \(32\) m
3. \(16\) m
4. \(4\) m

A particle is thrown with any velocity vertically upward, the distance travelled by the particle in the first second of its descent is:

1. g

2. $\frac{g}{2}$

3. $\frac{g}{4}$

4. Cannot be calculated

A body is thrown vertically upwards and takes 5 seconds to reach maximum height. The distance traveled by the body will be the same in

(1) 1^st and 10^th second

(2) 2^nd and 8^th second

(3) 4^th and 6^th second

(4) Both (2) & (3)

A ball is dropped from a bridge of 122.5 metre above a river. After the ball has been falling for two seconds, a second ball is thrown straight down after it. Initial velocity of the second ball so that both hit the water at the same time is

(1) 49 m/s

(2) 55.5 m/s

(3) 26.1 m/s

(4) 9.8 m/s

A balloon starts from the ground from rest with an upward acceleration of \(2\) m/s². After \(1\) sec, a stone is dropped from it. The time taken by the stone to strike the ground is approximately:

A boy throws balls into the air at a regular interval of 2 seconds. The next ball is thrown when the velocity of the first ball is zero. How high do the ball rise above his hand? [Take g = 9.8 m/ s²]

(1) 4.9 m

(2) 9.8 m

(3) 19.6 m

(4) 29.4 m

A ball projected from the ground vertically upward is at the same height at time t₁ and t₂. The speed of projection of ball is [Neglect the effect of air resistance]

(1) $g\left[t_2-t_1\right]$

(2) $\frac{g\left[t_1+t_2\right]}{2}$

(3) $\frac{g\left[t_2-t_1\right]}{2}$

(4) $g\left[t_1+t_2\right]$

Two balls are projected upward simultaneously with speeds of \(40\) m/s and \(60\) m/s. The relative position \((x)\) of the second ball with respect to the first ball at time \(t=5\) s will be: (neglect air resistance)
1. \(20\) m
2. \(80\) m
3. \(100\) m
4. \(120\) m