Recent content by helpinphysics

a ball is thrown horizontally from the roof a building 50m tall and lands 45m away from the base. what was the ball's initial speed?

i think i found the answer to the first part d=vit 5m=vi(2.7) 5/2.7=vi(2.7/2.7) 1.851=vi

for how long they are in the air i got 2.7s d=vit+(1/2)(9.81)t^2 35m=4.905t^2 7.13=t^2 t=2.67 rounded to 2.7s i can't find the speed

i have another quesstion...i did the second part right but I am not sure about the first. cliff divers push of horizontalll from rock platforms about 35m above the water, but they must clear rocky outcrops at water level that extend out into the water 5m from the base of the cliff directly...

thank you!

i got the answer it is 7.02 x=0+4.5*1.56 x=7.02m

i found the answer! x=0+4.5*1.56 x=7.02

so how do i apply that to the horizontal part

t=1.56s

but i guess a is 9.81m/s^2

i tried to use the equation but this is how i came up with my answer d=vit+(1/2)at^2 12=4.5t+(1/2)0t^2 12=4.5m/s*t t=2.66s

i am really stuck. i don't no what to next

ok i got that the time is 2.66s

what do i use to get the horizontal part

i don't think i did this right but is a .00849? i did 12 m=4.5m/s*2.66s+(1/2)at^2 12=11.97+3.53a .03/3.53=3.53a/3.53 a=.00849