Projection Under uniform Acceleration

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Question

The coach throws a base ball to a player with an initial speed of 20 ms^–1 at an angle of 45^o with the horizontal. At the moment the ball is thrown, the player is 50 m from coach. The speed and the direction that the player has to run to catch the ball at the same height at which it was released in ms^–1 is

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Solution

Range = $=\frac{(20)^2(1)}{10}=40\;m$
time $=\frac{2X20X\frac{1}{\sqrt2}}{10}=2\sqrt2\;sec$
Distance = 10 m
Hence ν = $\large \frac {10}{2\sqrt2}=\frac{5}{\sqrt2}\;m/s\;towards\;the\;coach$

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Projection Under uniform Acceleration

Remember concepts with our Masterclasses.

Range = time Distance = 10 mHence ν =

Ready to Test Your Skills?

free study material

Range = $=\frac{(20)^2(1)}{10}=40\;m$
time $=\frac{2X20X\frac{1}{\sqrt2}}{10}=2\sqrt2\;sec$
Distance = 10 m
Hence ν = $\large \frac {10}{2\sqrt2}=\frac{5}{\sqrt2}\;m/s\;towards\;the\;coach$