a ball is thrown straight up with an initial velocity of 20.0ft/s from a deck th

Question

a ball is thrown straight up with an initial velocity of 20.0ft/s from a deck that is 30.0ft. above ground.a)how long does it take to reach it's max. height?b)what max. height above the deck does it reach? c) at what speed does it hit the ground? d) what total length of time is the ball in the air?

Explanation / Answer

a) initial velocity, u = 20 ft/s acceleration due to gravity, g = 32 ft/s2 at max height, h (above deck) , velocity v = 0 ft/s v = u - gt => 0 = 20 - 32t (where, t sec is time taken to reach max. ht.) => t = 20/32 = 5/8 = 0.625 (ANSWER) b) h = ut - 0.5gt^2 = 20*5/8 - 0.5*32*(5/8)^2 = 6.25 ft (ANSWER) c) max. ht above ground = h+30 = 6.25+30 = 36.25 ft from this ht it starts falling downwards with initial velocity, u = 0 ft/s final velocity when it hits the ground, v= gt = 32t ft/s (t is time to reach ground) distance travelled, s = 36.25 = gt^2 = 32t^2 => t = sqrt(36.25/32) = 1.064 sec v = 32*1.064 = 34.1 ft (ANSWER) d) total time the ball in air = 0.625 + 1.064 = 1.689 sec (ANSWER)

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