A ball is hit upward at an angle theta from the horizontal and speed V = 30m/s.

Question

A ball is hit upward at an angle theta from the horizontal and speed V = 30m/s. It reactnes a maximum height of 15 m as shown in Fig, 1 Assume the trajectory start at ground level and ignore air resistance. (Ch. 3, pp.80-81) What is the initial angle theta ? how long is the ball in the air? How far is the ball from the starting point when it hits the ground ? How much work does gravity do on the ball when it reaches its peak height if it mass is 300 g ? What is the total work done on the ball by gravity ?

Explanation / Answer

a) Use equation,

T= F x L= F*L*sin = 15.5*0.255*sin33 = 2.15 N.m

b) Torque will be maximum if =90

Thus

Tmax = F x L = F*L*sin = 15.5*0.255*sin90 = 3.95 N.m

a) Use equation,

vfy^2 = viy^2 - 2gH

vfy^2 = (visin) -2*g*Hmax

at Hmax vfy=0m/s

0^2= (visin)^2 -2*g*Hmax

= sin^-1[sqrt[2*g*Hmax]/vi = = sin^-1[sqrt(2*9.8*15)/30] = 34.86 = 35 deg

b)

Use equation to calculate time for Hmax

y = viy*t-1/2gt^2

Hmax = visin*t-1/2gt^2

15 = 30sin34.86*t -1/2*9.8*t^2    => t= 1.75s

Time of flight = T=2t = 2*1.7 = 3.5s

c) x = vix*T = vicos34.86*t = 30*cos34.86*3.5 = 86.2 m

d) Wg = Fg*Hmax*sin = mg*Hmax*cos180 = 0.300*9.8*15*-1 = - 44.1 J

e) Wtotal = Fg*Hmax*sin = mg*x*cos90 = 0 J

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