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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

Emily Abbate

#8.14"m/s"#

First, the vertical component provides us with the flight time:

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

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

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

#t=2.21"s"#

Since the velocity's horizontal component is constant,

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

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Answer 2

William Audy

see equation [5] and use your calculator :D

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

[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 range of what you or your teacher uses determines the value of g [9,10].

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Answer 3

Layla Ahmed

The initial speed of the ball is approximately 12 m/s.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.