If the man on the left pulls on the object with a force of 500 N and the man on the right pulls on the object with a force of 75
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(a) The horizontal and vertical components of the ball’s initial velocity is 37.8 m/s and 12.14 m/s respectively.
(b) The maximum height above the ground reached by the ball is 8.6 m.
(c) The distance off course the ball would be carried is 0.38 m.
(d) The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
<h3>Horizontal and vertical components of the ball's velocity</h3>Vx = Vcosθ
Vx = 39.7 x cos(17.8)
Vx = 37.8 m/s
Vy = Vsin(θ)
Vy = 39.7 x sin(17.8)
Vy = 12.14 m/s
<h3>Maximum height reached by the ball</h3>Maximum height above ground = 7.51 + 1.09 = 8.6 m
<h3>Distance off course after 2 second </h3>Upward speed of the ball after 2 seconds, V = V₀y - gt
Vy = 12.14 - (2x 9.8)
Vy = - 7.46 m/s
Horizontal velocity will be constant = 37.8 m/s
Resultant speed of the ball after 2 seconds = √(Vy² + Vx²)
d = Δvt = (38.72 - 38.53) x 2
d = 0.38 m
The ball's velocity after 2.0 seconds if there is no crosswind is 38.53 m/s.
Learn more about projectiles here: brainly.com/question/11049671
The average velocity of Sandy is given by the total distance covered S divided by the total time taken t:
The total distance covered is
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
Therefore, the average velocity is
The total distance covered is
while the total time taken is 2 hours + half an hour (for the rest) + 1 hour and half, so
Therefore, the average velocity is