A ball player hits a home run and the baseball just clears a wall 8.00 meters high located at 106.2 meters from home plate. The ball is hit at an angle of 36.1 degrees to the horizontal, and air resistance is negligible. Assume the ball is hit at a height of 1.1 meters about the ground. The acceleration of gravity is 9.81m/s/s
1. What is the initial speed of the ball? (m/s)
2. How much time does it take for the ball to reach the wall? (seconds)
3. Find the speed of the ball when it reaches the wall. (m/s)
2007-11-26 13:16:55 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
y(t)=1.1+v0*sin(36.1)*t-.5*g*t^2
x(t)=v0*cos(36.1)*t
v0=106.2/(t*cos(36.1))
0=1.1-8+106.2*tan(36.1)-.5*9.81*t^2
t=3.79 sec
v0=34.66 m/s
3)vx=v0*cos(36.1)
vx=28 m/s
vy=v0*sin(36.1)-g*t
vy=-16.78 m/s
magnitude =32.65 m/s
j
2007-11-27 04:45:27 · answer #1 · answered by odu83 7 · 0⤊ 0⤋