y = y0 + vy0*t -1/2*g*t^2
x = x0 + vx0*t
where y is the y position as a function of time, y0 is the initial y position, vy0 is the initial y velocity, g is the acceleration due to gravity (9.8 m/s^2), t is time, x is the x position as a function of time, vx0 is the initial x velocity. Plugging in the numbers for when the ball clears the wall:
20.6 = 1.1 + v0*sin(35)*t - 1/2*9.8*t^2
132 = 0 + v0*cos(35)*t
Use algebra to solve this system of 2 equations with 2 unkowns: the time t, and the velocity v0.
The components of the velocity (vy = component, vx = x component) at this time are given by:
vy = dy/dt = v0y - g*t = v0*sin(35) - g*t
vx = dx/dt = v0x = v0*cos(35)
Solve for vy and vx at the time you found in part b
A ball player hits a home run, and the baseball just clears a wall 20.6 m high located 132.0 m from home plate. The ball is hit an an angle of 35° to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.1 m above the ground.
(a) What is the initial speed?
(b) How much time does it take for the ball to reach the wall?
(c) Find the components of the velocity and the speed of the ball when it reaches the wall
1 answer
1 answer