A ball is kicked from a location 7, 0, 6 (on the ground) with initial velocity 1

Question

A ball is kicked from a location 7, 0, 6 (on the ground) with initial velocity 10, 17, 2 m/s. The ball's speed is low enough that air resistance is negligible.

(a) What is the velocity of the ball 0.5 seconds after being kicked? (Use the Momentum Principle!)

(c) What is the average velocity of the ball over this time interval?

(d) Use the average velocity to find the location of the ball 0.5 seconds after being kicked.

Now consider a different time interval: the interval between the initial kick and the moment when the ball reaches its highest point. We want to find how long it takes for the ball to reach this point, and how high the ball goes.

(e) What is the y-component of the ball's velocity at the instant when the ball reaches its highest point (the end of this time interval)?

vyf =  m/s

(f) Fill in the missing numbers in the equation below (update form of the Momentum Principle: mvyf = mvyi + Fnet, yt).

m ___ = m ____ + mgt

(g) How long does it take for the ball to reach its highest point?

t =  s

(h) Knowing this time, first find the y-component of the average velocity during this time interval, then use it to find the maximum height attained by the ball.

ymax =  m

Explanation / Answer

Given data:
location(7,0,-6)
initial velocity (-19,17,-2)
time =0.5 seconds
a) after 0.5 s velocity of the ball
Vy = Vo + at = 17 - 9.8 * 0.5=12.1 m/s

c) the avg. velocity of the ball over this time interval is = (17+12.1)/2 = 14.55 m/s
and the average velocity of the ball is
Vavg = sqrt( (-10)^2 + (14.55)^2 + (-2)^2) m/s = 17.76 m/s
Then after 0.5s the location is <7-10*0.5, 0 + 14.55*0.5, -6 - 2*0.5>
which is < 2, 7.275, -7>

y component at max height = 0.
m * 0 = m * 17m/s + m * g * t
-17m/s = -9.8m/s * t
t = 1.7 s

Vavg,y = 17m/s / 2 = 8.5 m/s

max height = Vavg * t = 8.5m/s * 1.7s = 14.7 m

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