What is the horizontal component of the force on the ball after it leaves the throwers hands

Physics

1 answer:

e-lub [12.9K]3 years ago

5 0

After the ball leaves the thrower's hand, the only force on it is
due to gravity. There's no horizontal force acting on it at all. (C)

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A 1.0 kg football is given an initial velocity at ground level of 20.0 m/s [37 above horizontal]. It gets blocked just after re

stepan [7]

Refer to the diagram shown below.

Neglect air resistance.
The horizontal component of the launch velocity is
(20 m/s)*cos(37°) = 15.973 m/s
The vertical component of the launch velocity is
(20 m/s)*sin(37°) = 12.036 m/s

The acceleration due to gravity is g =9.8 m/s².
The time, t s, for the ball to reach a height of 3 m is given by
(12.036 m/s)*(t s) - (1/2)*(9.8 m/s²)*(t s)² = (3 m)
12.036t - 4.9t² - 3 = 0
2.4543t - t² - 0.6122 = 0
t² - 2.4563t + 0.6122 = 0
Solve with the quadratic formula.
t = (1/2)[2.4563 +/- √(6.0334 - 2.4488)]
t = 2.1748 or 0.2815 s
The ball reaches a height of 3 m twice.
The first time it reaches 3 m height is 0.2815 s.

Part a.
The vertical velocity when t = 0.2815 s is
Vy = 12.036 - 9.8*0.2815
= 9.2773 m/s
The horizontal component of velocity is Vx = 15.973 m/s
The resultant velocity is
√(9.2773² + 15.973² ) = 18.47 m/s
Answer:
The velocity at a height of 3.0 m is 18.5 m/s (nearest tenth)

Part b.
The horizontal distance traveled is
d = (15.973 m/s)*(0.2815 s) = 4.4964 m
Answer:
The horizontal distance traveled is 4.5 m (nearest tenth)