Ask a New Question

Question

The ball is thrown with an initial speed of 3.7 m/s at an angle of 15.4° below the horizontal. It is released 0.83 m above the floor. What horizontal distance does the ball cover before bouncing?
14 years ago

Answers

drwls
(horizontal velocity component) x (time to fall) =
(3.7 cos 15.4)* T

Get T by solving the equation
3.7 sin 15.4*T + (g/2) T^2 = 0.83

Take the positive root.
14 years ago

Related Questions

a golf ball is hit with an initial speed of 35 m/s at an angle less than 45 degress above the horizo... A golf ball is hit with an initial velocity of 50 m/s at an angle of 45 degrees with the horizontal.... A golf ball is hit with an initial velocity of 135 feet per second at an angle of 22 degrees above t... A golf ball is hit with an initial velocity of 50 m/s at an angle of 45° above the horizontal. What'... A stone is thrown with an initial speed of 15 m/s at an angle of 37 above the horizontal from the t... A golf ball is hit with an initial angle of 30.5° with respect to the horizontal and an initial velo... A ball is dropped from the initial height h1 above the concrete floor and rebounds to a height of eq... A golf ball is hit with an initial velocity of 140 ft per second at an inclination of 45 degrees to... A tennis ball is hit at an initial upward velocity of 25 feet per second. The situation can be mode... A tennis ball is hit at an initial upward velocity of 18 feet per second. The situation can be mode...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use