Advertisements
Advertisements
Question
A pebble is thrown vertically upwards with a speed of 20 m s-1. How high will it be after 2 s? (Take g = 10 m s-2)
Advertisements
Solution
Initial velocity, u = 20 m/s
Time, t = 2s
g = 10 m/s2
Maximum height reached in 2s, H = (1/2) gt2
Or, H = (1/2) (10) (2) 2
Or, H = 20 m
APPEARS IN
RELATED QUESTIONS
A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the
(a) force on the floor by the crew and passengers,
(b) action of the rotor of the helicopter on the surrounding air,
(c) force on the helicopter due to the surrounding air.
A constant force F = m2g/2 is applied on the block of mass m1 as shown in the following figure. The string and the pulley are light and the surface of the table is smooth. Find the acceleration of m1.
In the following figure, m1 = 5 kg, m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.
A body of mass 5 kg is moving with velocity 2 m s-1. Calculate its linear momentum.
Calculate the magnitude of force which when applied on a body of mass 0.5 kg produces an acceleration of 5 m s-2.
Prove mathematically F = ma
Multiple Choice Question. Select the correct option.
Which of the following are vector quantities?
A ball is thrown vertically downward with an initial velocity of 10 m/s. What is its speed 1 s later and 2 s later?
In the previous problem (5.3), the magnitude of the momentum transferred during the hit is ______.
A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 ms−1, q = 4 ms−2 and r = 5 ms−3. The force acting on the body at t = 2 seconds is ______.
