# Physics

A player throws a ball at an initial velocity of 36 m/s. The maximum distance the ball can reach (assume ball is caught at the same height at which it was released) is?
(I tried to solve it but i think it doesn't have all the information needed to slove it, but i'm not sure)

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1. I think i got it!
V^2/a
36^2/9.8
=1296/9.8
=132.2448
Rounding off;
=132m

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2. Vo = 36 m/s. @ What angle?.
We need the angle.

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3. but yass that v^2/a is equal to t(time) .. how can u relate it directly with
S( distance)??

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4. ok i think i got it...they asked for the maximum ditance... its mean maximum distance = range
as we know
R = V^2 sin2 theta/ g
v = 36m/s
= 36^2 sin 2(45) as range is maximum at 45 degree
= 1296/9.8
=132.2m = 132m approx

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5. 9908

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6. s=vi2/2g
s=(3362/229.8
s=1296/19.6
s=66.122 is reach the maximum distance but here we find out releace same position so distance double
2266.122
s=132 m

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7. Maximum distance=?
Maximum range angle=45
As we know that;
R=v^2sin2theta/g
R=36^2sin2 (45)/g
R=1296sin90/g
R=1296 (1)/g
R=1296/9.8
R=132m

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8. Data: Vo = 36 m/s. , for maximum distance or range always we use 45 So R = Vo sin 2 (45) / 9.8 Thus R = (36)square sin 90/ 9.8 Sin 90 = 1 so we have.R = (1296) (1) /9.8 we get. R = 132.24 m

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9. T=Vf-Vi ÷g
T=0-36 ÷9.8
T=3.67s
Now
X=Vox × t
X=36 × 3.67
X=132.24meter

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10. 9.jpg

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