Calculating Velocity of Ball at Impact with Ground

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Trandall
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Homework Statement



A ball is thrown at a velocity of 10 m/s, at an angle of 30° from horizontal and released 2 m
above the ground. What will the magnitude of ball’s resultant velocity be at the moment of
impact with the ground?

Homework Equations


Vf2 = vi2 + 2as


The Attempt at a Solution



Vf2 = 5(2) + 2 x -9.8 x -2
Vf2 = 25 + 39.2
Then find the square root

The answer is 11.8 m/s
What am I doing wrong, please help

Thanks
 
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Trandall said:

The Attempt at a Solution



Vf2 = 5(2) + 2 x -9.8 x -2
Vf2 = 25 + 39.2
Then find the square root

The answer is 11.8 m/s
What am I doing wrong, please help

Thanks

Split the velocity into components.

Use v2=u2-2g(s-s0) to find the vertical component of velocity when it hits the ground.
 
Im sorry, I am still stuck, could you write it out in the formulae, I am clueless when it comes to physics.
 
Trandall said:
Im sorry, I am still stuck, could you write it out in the formulae, I am clueless when it comes to physics.

v2=u2-2g(s-s0)

v= final velocity
u= initial velocity
g= acceleration due to gravity
s= displacement
s0=initial displacement
 
you'll have to analyze motions in the x & y direction separately as acceleration is along the y direction only.

X
[itex]{v}_{x0} = 10 \cos 30[/itex]
[itex]{a}_{x} = 0[/itex]
[itex]{v}_{x} = {v}_{x0}[/itex]

Y
[itex]{v}_{y0} = 10 \sin 30[/itex]
[itex]{a}_{y} = -g[/itex]
[itex]{s}_{y} = -2[/itex]

[itex]{s}_{y} = {v}_{y0}t + \frac{1}{2}{a}_{y}{t}^{2}[/itex]
solve for t & then use -
[itex] {v}_{y} = {v}_{y0} + {a}_{y}t[/itex]