Motion in a Straight Line

Motion in a Straight Line

Easy Questions
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Mechanics

Physical World, Units and Measurements
Motion in a Straight Line
Motion in a Plane
Laws of Motion
Work, Energy and Power
Gravitation
Mechanical Properties of Solids
Mechanical Properties of Fluids
System of Particles and Rotational Motion
  1. Two bodies, A (of mass 1 kg) and B (of mass 3 kg), are dropped from heights of 16m and 25m, respectively. The ratio of the time taken by them to reach the ground is​​​








  1. View Hint View Answer Discuss in Forum

    Let t1 & t2 be the time taken by A and
    B respectively to reach the ground then from the formula,

    h =
    1
    		gt²
    2

    For first body, 16 =
    1
    		gt1²
    2

    For second body, 25 =
    1
    		gt2²
    2

    ∴
    16
    		 =
    t1²
    		 ⇒
    t1
    		 =
    4
    25t2²t25

    Correct Option: C

    Let t1 & t2 be the time taken by A and
    B respectively to reach the ground then from the formula,

    h =
    1
    		gt²
    2

    For first body, 16 =
    1
    		gt1²
    2

    For second body, 25 =
    1
    		gt2²
    2

    ∴
    16
    		 =
    t1²
    		 ⇒
    t1
    		 =
    4
    25t2²t25

  1. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is​​​








  1. View Hint View Answer Discuss in Forum

    Let body takes T sec to reach maximum height.
    Then v = u – gT
    v = 0, at highest point. T = u/g …(1)
    Velocity attained by body in (T – t) sec v = u – g (T – t)

    = u – gT + gt = u – g
    u
    		 + gt
    g

    or v = gt …(2)

    ∴ Distance travelled in last t sec of its ascent
    s = (gt)t -
    1
    		 gt² =
    1
    		 gt²
    22

    Correct Option: C

    Let body takes T sec to reach maximum height.
    Then v = u – gT
    v = 0, at highest point. T = u/g …(1)
    Velocity attained by body in (T – t) sec v = u – g (T – t)

    = u – gT + gt = u – g
    u
    		 + gt
    g

    or v = gt …(2)

    ∴ Distance travelled in last t sec of its ascent
    s = (gt)t -
    1
    		 gt² =
    1
    		 gt²
    22

  1. ​A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so that more than two balls are in the sky at any time? [Given g = 9.8 m/s²]








  1. View Hint View Answer Discuss in Forum

    Let the required speed of throw be u ms–1. Then time taken to reach maximum height,
    t = u/g
    For two balls to remain in air at any time, t must be greater than 2.

    ∴
    u
    		 > 2 ⇒ u > 19.6 m/s
    g

    Correct Option: B

    Let the required speed of throw be u ms–1. Then time taken to reach maximum height,
    t = u/g
    For two balls to remain in air at any time, t must be greater than 2.

    ∴
    u
    		 > 2 ⇒ u > 19.6 m/s
    g

  1. If a ball is thrown vertically upwards with a velocity of 40 m/s, then velocity of the ball after two seconds will be (g = 10 m/s²)








  1. View Hint View Answer Discuss in Forum

    Initial velocity (u) = 40 m/s
    Acceleration a = –g m/s² = –10 m/s²
    Time = 2 seconds
    By Ist equation of motion,
    v = u + at
    v = 40 – 10 (2) = 20 m/s

    Correct Option: B

    Initial velocity (u) = 40 m/s
    Acceleration a = –g m/s² = –10 m/s²
    Time = 2 seconds
    By Ist equation of motion,
    v = u + at
    v = 40 – 10 (2) = 20 m/s

  1. A stone released with zero velocity from the top of a tower, reaches the ground in 4 sec. The height of the tower is (g = 10m/s²)








  1. View Hint View Answer Discuss in Forum

    Initial velocity (u) = 0; Time (t) = 4 sec and gravitational acceleration (g) = 10 m/s².
    Height of tower

    h = ut +
    1
    		 gt² = (0 × 4) +
    1
    		 10 × (4)²
    22

    Correct Option: C

    Initial velocity (u) = 0; Time (t) = 4 sec and gravitational acceleration (g) = 10 m/s².
    Height of tower

    h = ut +
    1
    		 gt² = (0 × 4) +
    1
    		 10 × (4)²
    22