Posted by hehe on .
) A batter hits a baseball so that it leaves the bat at speed
Vo = 37.0m/s at an angle ao = 53.1°, at a location where g =
9.80 m/S2. (a) Find the position of the ball, and the magnitude and direction of its velocity, at t = 2.00 s. (b) Find the time when the ball reaches the highest point of its flight and find its height h at this point. (c) Find the Iwrizontal range Rthat is, the horizontal distance from the starting point to where the ball hits the ground.

physics 
drwls,
(a) x = 37.0 cos53.1 * t = 22.22 t
y = 37.0 sin53.1 t  (g/2) t^2
= 29.59 t  4.90 t^2
Vx = 22.22
Vy = 29.59  9.8 t
angle to horizontal = arctan Vy/Vx
Plug in t = 2 s
(b) Compute y at time when Vy = 0
t = Vyo/g = 3.02 s
y = (Vyo)^2/(2g)
(c) R = 2 Vo^2 sin53.1*cos53.1/g
= 0.96 Vo^2/g
Vo is the launch velocity and Vyo is the y component of the launch velocity 
physics 
Jiban,
24.23m/s, 23.21 degree

physics 
yara,
Vo = 37 m/s
Theta = 53.1
Xo = 0 m i
Yo = 0 m j
ax = 0 m/s^2 i
ay = 9.8 m/s^2 j
Vox = Vo * costheta
= 37[cos(53.1)]
= +22.2 m/s i
Voy = Vo * sintheta
= 37[sin(53.1)]
Voy = +29.6 m/s j
x1 = +44.4 m i