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 = vixvix = 21 m s⁻¹

Acceleration = axax = 0 m s⁻²

Final velocity = vfxvfx = 21 m s⁻¹

Along Y-Axis

Initial velocity = viyviy = 0 m s⁻¹

Acceleration = ayay = g = 9.8 m s⁻²

Height = y = h = 10 m


To Find:

Horizontal distance = x = ?

Final velocity = vfvf = ?


Solution:

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

y = viyviy t + 1212 g t²  


by putting the corresponding values

10 m = 0 m s⁻¹ ｘ t + 1212 ｘ 9.8 m s⁻² ｘ t² 

10 m = 0 + 4.9 m s⁻² ｘ t² 

or

t² = 10m4.9ms⁻²


t² = √100m49ms⁻²


t = 107 s -------------(1)

Now we know that along x-axis

x = vix t

x = 21 m s⁻¹ ｘ 107 s

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

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

vfy = viy + ay t

putting the corresponding values

vfy = 0 m s⁻¹ + 9.8 m s⁻² ｘ 107 s

vfy = 14 m s⁻¹ ---------------  Ans. (2)

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

vf = √v2fx+v2fy

by putting the corresponding value

vf = √(21ms⁻¹)2+(14ms⁻¹)2

vf = √441m²s⁻²+196m²s⁻²

vf = √637m²s⁻²

vf = 25.13 m s⁻¹ -------------------Ans. (2)



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