Question

A ball thrown horizontally from a point 24 m above the ground, strikes the ground after traveling horizontally a distance of 18 m. With what speed was it thrown?

Answer 1

`8.14"m/s"`

Explanation:

First of all we get the time of flight from the vertical component:

`s=1/2"g"t^2`

`t=sqrt((2s)/(g))`

`t=sqrt((2xx24)/(9.8))`

`t=2.21"s"`

The horizontal component of velocity is constant so:

`v=s/t=18/2.21=8.14"m/s"`

Answer 2

see equation [5] and use your calculator :D

Explanation:

by setting the positive y-axis pointing upward and the positive x-axis pointing to the right,

[1] `y = y_0 + v_(0,y)*t -0.5*g*t^2`
with `y_0 = 24 m`

[2] `x = x_0 + v_(0,x)*t`
with `x_0 = 0`,

where
`v_(0,y) = 0` and
`v_(0,x) = v_0`

at where the ball strikes,
`18 m = x = v_0*t` then, `18 = v_0*t`
we then isolate t,
`t = (18 m)/(v_0)`

and insert t into (1) at the striking point, assuming that y = 0
[3] `0 = 24 + -0.5*g*((18 m)/(v_0))^2`
isolating `v_0`,
[4]`(v_0)^2 = (0.5*(18 m)^2*g)/(24)`
[5]`v_0 = sqrt((0.5*(18 m)^2*g)/(24))`

and the value of g depends on what you/your teacher uses ranging from [9,10].