Advertisements
Advertisements
Question
A player throws a ball upwards with an initial speed of 29.4 m s–1.
- What is the direction of acceleration during the upward motion of the ball?
- What are the velocity and acceleration of the ball at the highest point of its motion?
- Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.
- To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
Advertisements
Solution
- Since gravity is at work while the ball is in motion, the direction of gravity-related acceleration is always vertically downward.
- At maximum height, the velocity of the ball becomes zero. Acceleration due to gravity at a given place is constant and acts on the ball at all points (including the highest point) with a constant value, i.e., 9.8 m/s2.
- If we consider highest point of ball motion as x = 0, t = 0 and vertically downward direction as the positive direction of x-axis, then
- When moving upward, the position sign is negative, the velocity sign is negative, and the acceleration sign is positive, which means that v < 0 and a > 0.
- When moving downward, the position sign is positive, the velocity sign is positive, and the acceleration sign is also positive, which means that v > 0 and a > 0.
-
Initial velocity of the ball, u = 29.4 m/s
Final velocity of the ball, v = 0 (At maximum height, the velocity of the ball becomes zero.)
Acceleration, a = – g = – 9.8 m/s2
From the third equation of motion, height (s) can be calculated as:
`v^2- u^2 = 2gs`
`s = (v^2 - u^2)/2g`
`= ((0)^2 - (29.4)^2)/(2 xx (-9.8)) = 44.1 m`
From first equation of motion, time of ascent (t) is given as:
v = u + at
`t = (v - u)/a`
= `(-29.4)/-9.8`
= 3 s
Time of ascent = Time of descent
Hence, the total time taken by the ball to return to the player’s hands = 3 + 3 = 6 s
APPEARS IN
RELATED QUESTIONS
The following figure gives the x-t plot of a particle executing one-dimensional simple harmonic motion. Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
The velocity of a particle is towards west at an instant. Its acceleration is not towards west, not towards east, not towards north and towards south. Give an example of this type of motion .
A ball is projected vertically upward with a speed of 50 m/s. Find the time to reach the maximum height .
A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?
A stone is thrown vertically upward with a speed of 28 m/s.Find its velocity one second before it reaches the maximum height.
A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.
An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform?
An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.
A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find the horizontal distance it travels before reaching the ground .
A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find the maximum height reached .
A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?
Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed u at an angle θ with the horizontal.
A man is sitting on the shore of a river. He is in the line of 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.
A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. Find the time taken by the boat to reach the opposite bank.
A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river. How far from the point directly opposite to the starting point does the boat reach the opposite bank?
Suppose A and B in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of wind while maintaining the separation x. What will be the time B finds between seeing and hearing the drum beating by A?
A man is standing on top of a building 100 m high. He throws two balls vertically, one at t = 0 and other after a time interval (less than 2 seconds). The later ball is thrown at a velocity of half the first. The vertical gap between first and second ball is +15 m at t = 2 s. The gap is found to remain constant. Calculate the velocity with which the balls were thrown and the exact time interval between their throw.
