A ball is thrown horizontal from a height of 10 m with velocity of 21 m s⁻¹. How far off will it hit the ground and with what velocity? 

(Ans: 30 m,  25 m s⁻¹)



Given:

Along X-Axis

Initial velocity = `\v_{ix}` = 21 m s⁻¹

Acceleration = `\a_x` = 0 m s⁻²

Final velocity = `\v_{fx}` = 21 m s⁻¹

Along Y-Axis

Initial velocity = `\v_{iy}` = 0 m s⁻¹

Acceleration = `\a_y` = g = 9.8 m s⁻²

Height = y = h = 10 m


To Find:

Horizontal distance = x = ?

Final velocity = `\v_f` = ?


Solution:

We need find the time t first so using 2nd equation of motion along y-axis

y = `\v_{iy}` t + `\frac {1}{2}` g t²  


by putting the corresponding values

10 m = 0 m s⁻¹ ï½˜ t + `\frac {1}{2}` ｘ 9.8 m s⁻² ｘ t² 

10 m = 0 + 4.9 m s⁻² ｘ t² 

or

t² = `\frac {10 m}{4.9 m s⁻²}`


t² = `\sqrt frac {100 m}{49 m s⁻²}`


t = `\frac {10}{7}` s -------------(1)

Now we know that along x-axis

x = `\v_{ix}` t

x = 21 m s⁻¹ ｘ `\frac {10}{7}` s

x = 30 m ---------------Ans. (1)

Now using 1st equation of motion for projectile along y-axis

`\v_{fy}` = `\v_{iy}` + `\a_y` t

putting the corresponding values

`\v_{fy}` = 0 m s⁻¹ + 9.8 m s⁻² ï½˜ `\frac {10}{7}` s

`\v_{fy}` = 14 m s⁻¹ ---------------  Ans. (2)

The final speed vf at the time when it strike the ground is given by formula.

`\v_f` = `\sqrt {v_{fx}^2 + v_{fy}^2}`

by putting the corresponding value

`\v_f` = `\sqrt {(21 m s⁻¹)^2 + (14 m s⁻¹)^2}`

`\v_f` = `\sqrt {441 m² s⁻² + 196 m² s⁻²}`

`\v_f` = `\sqrt {637 m² s⁻²}`

`\v_f` = 25.13 m s⁻¹ -------------------Ans. (2)



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