A player throws a ball upward with an initial speed of 29.4 m s⁻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 the positive direction of the 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 m/s⁻² and neglect air resistance)
Answers
# Answers-
(a) Gravitational acceleration always acts in the downward direction towards the centre of the Earth whatever maybe the direction if motion.
(b) Velocity of the ball becomes zero at max height. Gravitational acceleration at a given place is constant and acts on the ball at all points with a constant value i.e. g=9.8m/s2.
(c) During upward motion, the sign of position is positive, sign of velocity is negative, and sign of acceleration is positive. During downward motion, the signs of position, velocity, and acceleration are all positive.
(d) Given here is-
u = 29.4 m/s
v = 0 (at max height)
a = –g = –9.8 m/s2
From 1st kinematic equation, time of ascent can be calculated as-
v = u + at
t = (v-u)/a
t = (0-29.4)/(-9.8)
t = 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.
Hope that solved your queries...
A) Since the ball is moving under the effect of gravity , the direction of accelration due to gravity will be in downwards direction which is vertically downwards
b) At maximum height, the final speed of all is zero and acceleration due to gravity will be acting vertically downwards.
c) When highest point is chosen as the location X=0 and t=0 and vertically downward
downward direction to be positive direction of x axis and upward direction is considered as negative direction.
During upward motion:
Position sign= negative
velocity=negative
acceleration= positive
During Downward motion :
Position= positive
sign of acceleration = positive
d) Let t be the time taken by ball reach maximum height
u=- 29. 4 m/s
a=9.8 m/s2
v=o
s=5m
t= 2 s
v²-u²=2as
0-(29.4 )² = 2x9.8xs
s=- (29.4)²/2x9.8
= - 44.4m
here negative sign shows distance covered is in upward direction
v=u+at
0=-29.4 +9.8 t
t-29.4/9.8
=3s
As time f ascent = time descent when object is free fall .
Total time= 3+3=6s