A cricketer strikes the cricket ball. The ball leaves the bat horizontally at 20 m/s. It hits the ground at a distance of 11 m from the point where it was struck. Assume that air resistance is negligible. (a) Name the type of motion exhibited by the ball (b) Calculate the time of flight of the ball. (c) Calculate the vertical velocity of the ball as it reaches the ground.
Answers
Answer :
(A) The type of motion exhibited by the ball is projectile motion.
(B) t = 0.55 s
(C) The vertical velocity of the ball as it reaches the ground is 5.39 m/s.
Explanation :
(a) The type of motion exhibited by the ball is projectile motion.
(b) To calculate the time of flight of the ball, we can use the following equation:
d = v_x * t
where d is the horizontal distance traveled by the ball, v_x is the horizontal velocity of the ball (20 m/s), and t is the time of flight.
So, substituting the values, we get:
11 = 20 * t
t = 11 / 20
t = 0.55 s
(c) To calculate the vertical velocity of the ball as it reaches the ground, we can use the following equation:
v_f = v_0 + g * t
where v_f is the final vertical velocity of the ball, v_0 is the initial vertical velocity (0 m/s), g is the acceleration due to gravity (9.8 m/s^2), and t is the time of flight.
So, substituting the values, we get:
v_f = 0 + 9.8 * 0.55
v_f = 5.39 m/s
Therefore, the vertical velocity of the ball as it reaches the ground is 5.39 m/s.
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