##### main algebraic solution

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May 29th, 2015
V is the given velocity at which the ball is struck
s = ut + ½gt²
calculate time needed for the ball to come back to rest using vertical components

s = displacement. so s will be zero when the ball comes back to rest(ie using the vertical components)
u = initial vertical velocity component
and t is time

u = VSinϴ = 60Sin30∘= 30

s = ut + ½gt²
0 = 30t + ½gt²
½gt² + 30t = 0
gt² + 60t = 0
t(gt + 60) = 0
t = 0, t = -60/g but initially the ball is going upwards so g is minus

time t = 0 the ball is initially at rest and time t = -60/g after travelling for time -60/g the ball comes back to rest.

(assume g = 10m/s² or 9.8m/s²)
i assume g = 9.8m/s²

so t = -60/g = -60/-9.8

Resolve the horizontal velocity component (ie along the X axis)
v = VCosϴ = 60Cos30∘

how far the ball travels horizontally = speed x time
d = vt = ( 60Cos30∘)(-60/-9.8 ) = 318.1m

2) same concept
u = 30Sin60

s = ut + ½gt²
(30Sin60)t + ½gt² = 0

t (30Sin60 + ½gt) = 0
t = 0 and t = -60Sin60 /g

v = 30Cos60 = 15m/s

d = vt = (15)(-60Sin60 /g) = 79.5m

May 29th, 2015

its suppose to be in feet....not meters

May 29th, 2015

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May 29th, 2015
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May 29th, 2015
Oct 24th, 2017
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