Launch a tennis ball at 20.0 m/s at 60 degrees above horizontal from the top of

Question

Launch a tennis ball at 20.0 m/s at 60 degrees above horizontal from the top of a building that is 50.0 meters high. My issues are with (e) and (g). The others I feel pretty good about but feel free to point out any errors.

a) find the initial horizontal component of the velocity.

b) find the vertical component of the velocity.

c) find the maximum height, measuring from ground level.

d) find the time to reach the maximum height.

e) find the magnitude of the vertical velocity component when it hits the ground.

This I don't have. Since after 1.77 sec the ball is falling under the force of gravity I want to take the total time from a later question and do this: (5.42 sec - 1.77 sec)(9.8 m/s^2)=35.8 m/s. Would this be correct?

f) find the speed of the tennis ball when it strikes the ground.

g) find the direction of the velocity when it strikes the ground.

If I am thinking about (e) correctly than I want to do tan^-1(35.8/10.0)=74.4 degrees. This doesn't seem right to me. I would think that angle should be in the fourth quadrant.

h) find the total time in the air.

i) find the horizontal distance when it strikes the ground, measuring from the base of the building.

j) find the speed of the ball 10 meters above the ground.

I included everyting just in case it is needed to find (e) and (g). I tried to just load the worksheet itself but I guess the file was to large. Please help.

Explanation / Answer

a) Vox = Vo*os60 = 10 m/s

b) voy = Vo*sin60 = 17.3 m/s

c) y - yo = voy^2/2g = 15.3

y = 65.3 m

d) t = voy/g = 1.77 s

e) a = -g         y = -50 m

vfy ^2 - viy^2 = 2*a*y

vfy^2 - 17.3^2 = 2*9.8*50

vfy = 35.8 m/s

your method is also correct

f) vf = sqrt ( vfy^2 + vox^2)

vf = 37.2 m/s

g)

vfy is in down wards diretion

tan^-1(-35.8/10) = -74.4

h) 5.42s

i) D = 10*5.42 = 54.2

j) Ei = Ef

KEi + PEi = KEf + PEf

0.5*m*vi^2 + m*g*h1= 0.5*m*vf^2 + m*g*h2

20^2 + 2*9.8*50 = vf^2 + 2*9.8*10

vf = 34.4 m/s

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