a. X = 1.30m/s * 0.25s = 0.325 m.
Y = yo - 0.5g*t^2.
Y = 1.25 - 4.9*(0.25)^2 = 0.944 m. above gnd.
b. X = 1.30m/s * 0.500s =
Y = 1.25 - 4.9*(0.50)^2 =
c. Vy = Vo + g*t.
Vy = 0 + 9.8*0.5 = 4.9 m/s. = Ver. component of velocity.
Vx = 1.30 m/s and remains constant.
V = Sqrt(Vx^2+Vy^2)
Tan A = Vy/Vx = 4.9/1.30 =
A = ?.
A person skateboarding with a constant speed of 1.30 m/s releases a ball from a height of 1.25 m above the
ground. Given that x0 = 0 and y0 = h = 1.25 m, find x and y for (a) t = 0.250 s and (b) t = 0.500 s. (c) Find the velocity,
speed and direction of motion of the ball at t = 0.500 s.
1 answer
1 answer