Vo = 25m/s[20o].
Xo = 25*Cos20 = 23.5 m/s. = hor. component.
Yo = 25*sin20 = 8.6 m/s = ver. component.
a. Xo * T = 50.
23.5*T = 50,
T = 50/23.5 = 2.13 s. = time in air.
Note: 50 is suppose to be max. range, but that is an error.
Range = Vo^2*sin(2A)/g.
Range = 25^2 * sin(40)/9.81 = 41 m., max. But we used what they gave us.
b. Y^2 = Yo^2 + 2g*h = 0.
8.6^2 - (19.6)h = 0,
h = ?
tr
A baseball is thrown at an angle of 20º relative to the ground at a speed of 25.0 m/s. If the ball was caught 50.0 m from the thrower, how long was it in the air?How high did the baseball ravel before beginning it's descent?
Please help me!!!
1 answer
1 answer