A baseball is hit into the air at an initial speed of 35.5m/s at an angle of 49.9 degrees above the horizontal. At the same time the center fielder starts running away from the batter and catches the ball 0.877m above the level at which it was hit. If the center fielder is initially 116m from home plate, what must be his average speed?
physics - bobpursley, Friday, September 12, 2008 at 8:54pm
You have two dimensions to work with: horizontal and vertical
hfinal=hinitial+35.5sin49.9*t -1/2 g t^2
let hinitial be zero, so hfinal is .877.
You can solve for time here, using the quadratic equation.
distance=35.5cos49.9 t solve the is for distance the ball went.
Finally, the distance the ball player ran is the above -116.
physics - GK, Friday, September 12, 2008 at 9:04pm
The initial vertical component of the velocity is:
Vyo = (35.5m/s)(sin49.9) = _____ m/s
The vertical distance can be described by:
y = (Vyo)t + (1/2)(-9.8m/s^2)t^2
At the time the ball was caught:
0.877m = (Vyo)t + (1/2)(-9.8m/s^2)t^2
Substitute the value for Vyo and rearrange into a quadratic equation in standard quadratic trinomial form.
Solve for the time, t and use the larger of the two values.
The center fielder’s speed is (116m) / t = _____ m/s
physics - bobpursley, Friday, September 12, 2008 at 9:26pm