Question

Launch a tennis ball at 20.0 m/sec at 60 degrees above horizontal from the top of a building that is 50.0 meters high. Please show all your work on this page. Keep answers in three significant figures.

a. Find the initial horizontal component of the velocity

b. Find the vertical component of the velocity

c. Find the maximum height, measuring from the ground level

d. Find the time to reach the maximum height

e. Find the magnitude of vertical velocity component when it strikes the ground

.

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

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

Answer 1

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