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?

2 Answers
Michael
Nov 2, 2015

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"

Donatello N.
Nov 2, 2015

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].

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