Refer to the conceptual example 3-9 on page 67 of your textbook. Suppose the boy

Question

Refer to the conceptual example 3-9 on page 67 of your textbook. Suppose the boy in the tree is 25 feet above the ground before dropping. The other boy set up on the hilltop has a ball shooter that will launch a projectile at a velocity of 25 m/s. At the instant the boy on the hilltop shoots the ball, the other boy releases his grip on the tree branch and drops downward. Parts (a), (b), (c) one point each, part (d) two points. Convert the height of the boy in the tree to meters. Do the kinematic equations we derived in chapter 2 apply in this situation? Explain. Express the initial velocity of the launched ball in vector notation. Under these conditions, what distance d (shown on the diagram as the x-axis horizontal distance) does the boy on the hill need to be from the boy in the tree in order to hit him with the ball when the boy is still 1 meter above the ground?

Explanation / Answer

a) height ,H = 25 ft

1ft = 0.3048 m

H= 25*0.3048 = 7.62 m

b) In this situation we can apply kinematic equations.

since boy falling from tree has initial velocity,u = 0

and will fall due to acceleration due to gravity ,a= g

distance travelled ,s= h = 7.62 m

so from V2 - u2 = 2 aS

we can calculate the final velocity ,V.Hence we can apply equations of kinematics.

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c) initial velocity of ball in vector form, V = 25 i + 0j

in Y- direction initial velocity = 0 that's why 0j.

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d) since diagram is not given ,i'm solving this according to data given.

in horizontal direction , Sx= ux*t + ( 1/2)*ax*t2

sx= R ( horizontal distance)

ux= 25 m/s

ax= 0 ( in horizontal direction velocity = consatnt)

so R = 25* t -------------------------1

in Y- direction , Sy = uy*t + (1/2)*ay*t2

Sy = vertical distance = 7.62- 1 = 6.62 m

ay = g

uy = 0 ( initial velocity in y- direction)

t = time

so 6.62 = 0 + ( 1/2)*9.8*t2

t = 1.16 sec

from eq 1, R= 25*1.16 = 29 m

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