A person throws a ball vertically upward, in the positive y-direction, with an i

Question

A person throws a ball vertically upward, in the positive y-direction, with an initial speed v0y = 29.4 m/s. The release point is at y0 = 0.00 m. After release, the (constant) acceleration of the ball due to gravity is ay = 9.80 m/s2 . The vertical position of the ball versus time is designated as y(t). In this problem ignore the possible influence of air resistance on the motion of the ball.

(a) Write down an expression for the position y of the ball versus time.

(b) Write down an expression for the velocity vy of the ball versus time. How long does it take for the ball to reach its maximum height?

(c) What is the maximum height y of the ball before the ball starts coming back down?

(d) How long does it take for the ball to reach the initial height y = y0 again after it is launched?

(e) What are the velocity and speed of the ball when the ball returns to y = y0 = 0.00 m?

Explanation / Answer

a) y = yo + voy*t + (1/2)*ay*t^2

b) vy = voy + ay*t

let t_up is the time taken for the ball to reach maximum height.

at maximum height, vy = 0

so, use vy = voy + ay*t

0 = voy - g*t_up

==> t_up = voy/g

c) maximum height,

h = yo + voy*t_up + (1/2)*ay*t^2

= 0 + voy*(voy/g) - (1/2)*g*(voy/g)^2

= voy^2/(2*g)

d) total time flight, t_flight = 2*t_up

= 2*voy/g

e) velocity, vy = -voy

speed = voy

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