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=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.
• (d) To what height does the ball rise and after how long does the ball return to the player’s hands?(take g=9.8 ms^-2 and neglect air resistance)

Solution:

• (a) Vertically downwards.
• (b) At the highest point velocity =0 and acceleration a=-g=-9.8 ms^-2

Upward direction:

Position-Positive

Velocity-negative

Acceleration-positive

Downward direction:

Position-positive

velocity – positive

Acceleration-positive

• How long does the ball return to the player’s hand?

$${u=gt}$$

$${t=\frac {u}{g}}$$

$${t=\frac{29.4}{9.8}}$$

$${t= 3\sec}$$

Total time taken =Time taken to reach top +Time taken to reach the bottom

$${3 \ s +3 \ s =6 \ s}$$

• (d) To what height does the ball rise?

$${h=ut-\frac{1}{2}gt^2}$$

$${h=(29.4)(3)-\frac{1}{2}(9.8)(3^2)}$$

$${h=88.2-44.1}$$

$${h=44.1 \ m} $$