Motion in One Dimension Problem - 1

Description:

Question

A player throws a ball upwards with an initial speed of 29.4 ms^-1

a. What is the direction of acceleration during the upward motion of the ball?

b. What are the velocity and acceleration of the ball at the highest point of its motion?

c. Choose the x = 0m and t = 0s to be the location and time of the ball at its highest point. Vertically downward direction to be the positive direction of xaxis, and give the signs of position, velocity and acceleration of the ball during its upward and downward motion.

d. To what height does the ball rise and after how long does the ball return to the player’s hands?

Solution

a. Acceleration is given by −

a = change in velocity/time

a = V_final - V_initial/t

As the ball is thrown upwards velocity will keep on decreasing due to force of gravity, so

V_final > V_initial

Hence,

a = negative

As initial velocity is 29.4 ms^-1 (Positive), so upward direction is considered positive, hence negative acceleration indicates that direction of acceleration is downwards.

b. As mentioned above, as the ball is thrown upwards velocity will keep on decreasing due to force of gravity, so velocity at the highest point is zero.

Although the velocity is zero, acceleration due to gravity is applicable on the ball, hence at the highest point acceleration is 9.8 ms^-2 downwards.

c. Downward direction is assumed to be positive, so

For downward motion −

• Sign of position is positive
• Sign of velocity is positive
• Sign of acceleration is positive

For upward motion −

• Sign of position is positive
• Sign of velocity is negative
• Sign of acceleration is positive.

d. Consider the height to be h (displacement)

We have u = 29.4, a = 9.8(downwards)

At the highest point velocity is zero, hence final velocity is v = 0

By using kinematic equation −

h = v² - u²/2a

S =

0² - 29.4²/2(-9.8)

S = 44.1 m

Hence, ball will rise to a height of 44.1 m