A cannon ball is fired horizontally with a velocity of 50m/s from the top of...

A cannon ball is fired horizontally with a velocity of 50m/s from the top of a cliff 90m high.After how many seconds will it strike the plain at the foot of the cliff? At what distance from the foot of the cliff will it strike? With wat velocity will it strike the ground?

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Answers (2)

ROBERT
2 months ago
h= 90m
u= 50m/s
h= ut + ½gt²
90= 50t + ½(10)t²
90= 50t + 5t²
18= 10t + t²
t² + 10t - 18= 0
t= [-b ± √(b² - 4ac)]/ 2a
a= 1, b= 10; c= -18
t= [- 10 ± √(10² - 4(1)(-18)]/2(1)
= -10 ± √(100 + 72) / 2
= -10 ± √(172) / 2
= -10 + √172 / 2 or - 10 - √172 / 2
= 1.56 or -11.56
t= 1.56s

S= (u + v / 2) t
S= (50 + 0 / 2)(1.56)
= 25(1.56)
= 39m

v²= u² + 2gh
v²= 50² + 2(10)(90)
= 2500 + 1800
= 4300
v= √4300
= 65.57m/s
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