t. A goll bal in hit with initial velocity initial velooity 7sm\'sD a) How kong

Question

t. A goll bal in hit with initial velocity initial velooity 7sm'sD a) How kong is the ball in the alr? by What horizontal distance does travels er ball travet of 19a m/s at an angle of e3 up from the horiontal The ball twavels over and returne to a) What are the initial by For how kong is the fowotball in ) What horizontal distance doees the footbaill travel d) What maximum height does the foothall reach? the air? 3 A small frog jumps forward on level ground, landing 0600 s later a) With what vertical speed did the frog take off ? b) To what maximam vertical height does the frog jump? c) How far ahead of its starting point does the frog land if its takeoff is at 30.0 above the horizontall7 4. A basehall is tossed from one player to another who is 400 m away. I the ball is in the air for 200 s and is caught at the same level it was thrown, with what velocity was it thrown? 5. A long jumper takes off at an angle of 20.0P above the horizontal and reaches a maximum height of 60.0 cm at mid-flight a) What is his forward speed? b) How far forward does he jump? A golf ball is hit and lands on level ground 151 m away velocity was it hit? 6. 38s later. With what takes off from a ramp that is angled at 20.0° above the horizontal and over 14 school buses that are each 3.00 m wide. If the motorcycle just clears the jumps last bus and lands on a second ramp that is at the same level as the first. initial speed? swers: 1. a) 1.0s b) 6.1 m 2 a) 17.0 m/s [upl, 9.80 m/s [horizontal] b) 3.46s c) 33.9 m d) 14.7 m 5. a) 9.42 m/s b) 6.59 m 6. 52 m/s (22' above the horizontalj 7. 25.3 m/s

Explanation / Answer

1)

along vertical

y - y0 = voy*t + (1/2)*ay*t^2

0 - 0 = 7.8*sin39*t - (1/2)*9.8*t^2

t = 1 m/s

(b)

along horizontal

X = vox*t = v*costheta*t

X = 7.8*cos39*1

X = 6.1 m

============================

(2)

a)

initial vertical component of velocity = voy - vo*sintheta = 19.6*sin60 = 17 m/s

initial vertical component of velocity = vox - vo*costheta = 19.6*cos60 = 9.8 m/s

(b)

along vertical displacement y = 0

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

0 = 17*t - (1/2)*9.81*t^2

t = 3.46s

(c)

X = vox*t + (1/2)*a*t^2

a = 0

X = 9.8*3.46 = 3.9 m

(d)

maximum height Hmax = voy^2/(2*g) = 17^2/(2*9.81) = 14.7 m

==========================

3)

a)

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

0 = v0y*0.6 - (1/2)*9.81*0.6^2

voy = 2.94 m/s

b)

Hmax = voy^2/2g = 2.94^2/(2*9.8)= 0.441 m

(c)

x = vox*t = vo*cos30*t

x = (voy/sin30)*cos30*t

x = voy*cot30*t = 2.94*cot30*0.6 = 3.05 m

==================================

4)

along vertical

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

0 = voy*2 - (1/2)*9.8*2^2

voy = 9.8 ms/

along horzontal

x = vox*t

40 = vox*2

vox = 20 m/s

vo = sqrt(vo^2+voy^2) = 22.3 m/s

========================

5)

maximum height = voy^2/(2g) = (v*sintheta)^2/(2g)

0.6 = (v*sin20)^2/(2*9.81)

v = 10.03 m/s

forward speed vox = v*cos20 = 9.42 m/s

===============================

6)

time t = (2*voy)/g

3.8 = (2*v0y)/9.81

v0y = 19 m/s

x = vox*t

181 = vox*3.8

vox = 48 m/s

v0 = sqrt(vo^2+voy^2) = 52 m/s

============================

7)

time of flight = t = 2*v*sintheta/g

along horizontal

x = v*costheta*t

x = v*costheta*2*v*sintheta/g

x = v^2*sin(2theta)/g

14*3 = v^2*sin(2*20)/9.81

v = 25.3 m/s

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