A child throws a ball with an initial speed of 8.00 m/s at an angle of 40.0° above the horizontal. The ball leaves her hand 1.00 m above the ground. What is the magnitude of the ball's velocity just before it hits the ground?
A) 1.22 m/s
B) 6.79 m/s
C) 7.45 m/s
D) 9.14 m/s
E) 4.58 m/s

Question

Quizplus · Accepted Answer

The magnitude of the ball's velocity just before it hits the ground can be found using energy conservation or kinematic equations. The initial kinetic energy and potential energy are converted into kinetic energy just before impact. The velocity can be calculated using the formula: v = sqrt(vx^2 + vy^2), where vx = v0 * cos(θ) and vy = sqrt(v0^2 * sin^2(θ) + 2gh). Solving gives approximately 9.14 m/s.

A Child Throws a Ball with an Initial Speed of 8.00