A baseball is hit so that it travels straight upward after being struck by the b

Question

A baseball is hit so that it travels straight upward after being struck by the bat. A fan observes that it takes 3.20 s for the ball to reach its maximum height. (a) Find the ball's initial velocity. (b) Find the height it reaches. It is possible to shoot an arrow at a speed as high as 127 m/s. (a) If friction is neglected, how high would an arrow launched at this speed rise if shot straight up? (b) How long would the arrow be in the air? A certain automobile manufacturer claims that its deluxe sports car will accelerate from rest to a speed of 47.0 m/s in 7.15 s. (a) Determine the average acceleration of the car. (b) Assume that the car moves with constant acceleration. Find the distance the car travels in the first 7.15 s. (c) What is the speed of the car 10.0 s after it begins its motion if it continues to move with the same acceleration?

Explanation / Answer

8)

a)

using the kinematic eqn v = v0 -gt

Here v = 0 at the top.

So v0 = g*t = 9.8m/s^2*3.20s = 31.4m/s

b)

Use v^2 = v0^2 - 2*g*ymax

again v = 0

ymax = v0^2/(2*g) = (31.4m/s)^2 / (2*9.8m/s^2) = 50.2m

9)

Given

Vo = 127 m/s

g = 9.81 m/s^2

Theta = 90

Calculate vertical component of initial velocity

Vyo = Vo * Sin(theta)

Vyo = (127 m/s) * (1)

Vyo = 127 m/s

Calculate how long it will take it to rise to the top of its trajectory

Tr = Vyo / g

Tr = (127 m/s) / (9.81 m/s^2)

Tr = 12.95 s

Use that to calculate the maximum height reached (yo below is only used in situations where your launch point is

higher than where the object will land)

H = yo + (Vyo * t) - (0.5 * g* t^2)

H = (0 m) + [ (127 m/s)*(12.95 s) ] - [ 0.5*(9.81 m/s^2)*(12.95 s)^2 ]

H = (1644 m) - [ 823 ]

H = 821 ...............(A)

Find the total length of time by adding time of rise and fall

Tt = Tr + Tf

Tt = (12.95 s) + (12.95 s)

Tr = 25.9 s .................(B)

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