*I KNOW WE\'RE NOT SUPPOSE TO ASK MULTIPLE QUESTIONS IN ONE,BUT I\'M NOT UNDERST
ID: 1967680 • Letter: #
Question
*I KNOW WE'RE NOT SUPPOSE TO ASK MULTIPLE QUESTIONS IN ONE,BUT I'M NOT UNDERSTANDING,CAN SOMEBODY PLEASE HELP ANYWAY?? MAX POINTS REWARDED!! THANX*
A ball is thrown from a high cliff at an angle of -30degrees from the horizontal with an initial velocity Vo=15m/s. It strikes the ground 3.0 seconds later.
(a) What is the distance from the base of the building that the ball strikes the ground?
(b)How high is the cliff?
(c)How long does it take for the ball to reach 10 meters below the cliff top?
(d) What is the velocity vectore before the ball hhits the ground?
*THANK YOU IF U ANSWER THIS!! MAX POINTS REWARDED!!!*
Explanation / Answer
x component of velocity does not change = vcos
therefore, horizontal displacement = distance from the base of the building to the point where the ball strikes ground = vcos x t = 15cos-30 x 3 v= 38.97 m
Svertical = height of building = (usin)t-1/2gt2 = 15(sin-30)x3 - 4.9x32 = 66.6m
-10 = (usin)t-1/2gt2= -7.5t-4.9t2, solve the quadratic equation to get t =0.8553 seconds
vvertical = uvertical + gt = 15sin(-30) - 9.8x3 = -36.9 m/s
vhorizontal is constant as said earlier = 15cos(-30) =12.99 m/s
magnitude of v =( 12.992+36.92) =39.12 m/s
it is at an angle tan-1(vvertical/vhorizontal) = -70.6o with horizontal
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