Aball reaches a height of 14 meters and lands 21 meters away. what is the initial velocity?

1. solve for y-dir

sy = v0yt - 0.5gt^2 where g = 9.81 m/s^2
14 = v0yt - 4.905t^2

vy = v0y - gt where vy = 0 at the top of height
0 = v0y - 9.81t

2 eqns, 2 unkns
i. 14 = v0yt - 4.905t^2
ii. 0 = v0y - 9.81t

solved
v0y = 1.689 m/s
t = 16.573 s

2. solve for x-dir

sx = 0.5t×(vx + v0x)
where vx = 0 m/s b/c it lands in the end
where t = 16.573×2 = 33.146 s b/c 16.573 seconds was until ball reached height, hence it must take twice as long to reach the ground

21 = 0.5×33.146*(0 + v0x)

v0x = 1.267 m/s

Now combine x & y components

v0 = sqrt(v0x^2 + v0y^2)

m/s

hence v0 = 2.112 m/s

The rearrangement of particles in a physical change. hope this

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