A ball is thrown horizontally from a height of #24 m# and hits the ground with a speed that is five times its initial speed. What is the initial speed?

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Answer 1 · 2018-04-02T15:03:31

#"4.43 m/s"#

Let the initial speed be #"u"#. This is along horizontal. This speed will remain constant throughout the journey.

Horizontal component of velocity when it hits the ground is

#"v"_"x" = "u"#

If #("v"_"y")# be the vertical component of velocity when it hits the ground then,

#"v"_"y"^2 = "2gH" = "2 × 9.8 m/s"^2 × "24 m" = 470.4\ "m"^2//"s"^2#

Final speed of object #= sqrt("v"_"x"^2 + "v"_"y"^2) = sqrt("u"^2 + 470.4)#

According to question

#sqrt("u"^2 + 470.4) = 5"u"#

Squaring on both the sides

#"u"^2 + 470.4 = "25u"^2#

#24"u"^2 = 470.4#

#"u"^2 = 470.4/24 = 19.6#

#"u" = 14/sqrt(10)\ "m/s" = "4.43 m/s"#