A player throws a ball upwards with an initial speed of 29.4 m s⁻¹: Choose the x = O m and t = O 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.
Answers
Answer: HIIII
Explanation:
# 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.
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