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
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"#

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].