From a 200 m high tower, one ball is thrown upwards with speed of 10 1ms^-1 and another is thrown vertically downwards at the same speeds simultaneously. The time difference of their reaching the ground will be nearest to

12 s
6 s
2 s
1 s

Solution

The ball thrown upward will lose velocity in 1 s. It return back to thrown point in another 1 s with the same velocity as second. Thus the difference will be 2 s.

A body travels 2 m in the first two second and 2.20 m in the next 4 second with uniform deceleration. The velocity of the body at the end of 9 second is

–10 1 ms^-1
– 0.201 ms^-1
– 0.40 1 ms^-1
– 0.801 ms^-1

Solution

If you were to throw a ball vertically upward with an initial velocity of 50 m/s,approximately how long would it take for the ball to return to your hand? Assume air resistance is negligible.

2.5 s
5.0 s
7.5 s
10 s

Solution

The only force acting on the ball is the force of gravity.The ball will ascend until gravity reduces its velocity to zero and then it will descend. Find the time it takes for the bal lto reach its maximum height and then double the time to cover the round trip.

Using v_{at maximum height}= v₀ + at = v₀– gt, we get:0 m/s = 50 m/s – (9.8 m/s²) t

Therefore,
t = (50m/s)/(9.8 m/s²) ~ (50 m/s)/ (10 m/s²) ~ 5s

This is the time it takes the ball to reach its maximum height. The total round trip time is 2t ~ 10s.

The motion of a particle is described by the equation u = at.The distance travelled by particle in first 4 sec is

Solution

Equation of motion is u=at

An automobile travelling with a speed of 60 km/h, can apply brake to stop within a distance of 20m. If the car is going twice as fast i.e., 120 km/h, the stopping distance will be

60 m
40 m
20 m
80 m

Solution

From a 10m high building a stone ‘A’ is dropped&simultaneously another stone ‘B’ is thrown horizontally with an initial speed of 5m/sec^–1. Which one of the following statements is true?

It is not possible to calculate which one of two stones will reach ground first
Both stones ‘A’ & ‘B’ will reach the ground simultaneously.
‘A’stones reach the ground earlier than ‘B’
‘B’ stones reach the ground earlier than ‘A’

Solution

Since in both case the height of building and downward acceleration ‘g’ is same. So both stones reach simultaneously i.e.,S=¹⁄₂ gt² ⇒ 10 = ¹⁄₂ 10 × t²

or t,=\(\sqrt{2}\)sec for both stone.

A stone is thrown vertically upwards. When the particle is at a height half of its maximum height, its speed is 10m/sec,then maximum height attained by particle is (g= 10m/sec²)

8 m
10 m
15 m
20 m

Solution

From third equation of motion

v² = u² – 2gh (∵a=-g)

Given, v =10 m/sec at h/2. But v =0, when particle attained maximum height h.

Therefore (10)² = u² – 2gh/2

or 100 = 2gh –2gh/2 (∵0 = u² – 2gh)

⇒ h = 10 m

The relative velocity V_AB or V_BA of two bodies A & B maybe

(1)greater than velocity of body A

(2)greater than velocity of body B

(3)less than the velocity of body A

(4)less than the velocity of body B

(1) and (2) only
(3) and (4) only
(1), (2) and (3) only
(1), (2), (3) and (4)

Solution

All options are correct :

(i) When two bodies A & B move in opposite directions then relative velocity between A & B either V_AB or V_BA both are greater than V_A & V_B.

(ii) When two bodies A & B move in parallel direction

then V_AB=V_A-V_B⇒V_AB < V_A

V_BA=V_B-V_A⇒V_BA < V_B

Similar balls are thrown vertically each with a velocity 20 ms^-1, one on the surface of earth and the other on the surface of moon. What will be ratio of the maximum heights attained by them? (Acceleration on moon = 1.7 ms^-2 approx)

6
¹⁄₆
¹⁄₅
None of these

Solution

A particle is moving in a straight line with initial velocity and uniform accelerationa. If the sum of the distance travelled in t^th and (t + 1)^th seconds is 100 cm, then its velocity after t seconds, in cm/s, is

Solution

The distance travel in n^th second is

S_n= u + ¹⁄₂ (2n–1)a....(1)

so distance travel in t^th & (t+1)^th second are

S_t= u +¹⁄₂ (2t–1)a....(2)

S_t+1= u+¹⁄₂ (2t+1)a....(3)

As per question,St + St + 1 = 100 =2 (u+ at)....(4)

Now from first equation of motion the velocity, of particle after time t, if it moves with an accleration a is

v = u + a t....(5)

where u is initial velocity

So from eq(4) and (5), we get v = 50cm./sec.

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