The catapult hurls a stone of mass 32.0 g with a velocity of 50.0 m/s at a 30.0' angle of elevation. (a) What is the maximum height reached by the stone? (b) What is its range? (c) How long has the stone been in the air when it returns to its original height?
(Ans. (a) 31.87 m (b) 5.1 s (c) 220.8m)


Given:

Initial velocity of the bullet = vi = 50 m s⁻¹  
Angle = 𝞱 = 30°
Value of g = 9.8 m s⁻²

To Find:

(a) Maximum height of stone reached = H = ?
(b) Time of flight of stone = T = ?
(c) Range of stone = R = ?


Solution:

(a) Maximum height of stone reached = H = ?

As the stone hurled from the catapult will follow the path of projectile motion so, the formula to find the height reached by a projectile is given; 

h = `\frac {vi² sin² θ}{2g}`


h = `\frac {(50 m s⁻¹ )² sin² 30}{2ｘ9.81 m s⁻² }`


h = `\frac {(2500 m² s⁻² ) (0.25)}{19.62 m s⁻² }`


h = 31.855 m


or


h = 31.8 m ----------------Ans.



Thus the maximum height reached by the stone is 31.9 m


(b) Time of flight of the stone = T = ?


The formula to find the total time of flight of a projectile is given;

T = `\frac {2vi sin θ}{g}`

T = `\frac {(2ｘ50 m s⁻¹ ) sin 30}{9.81 m s⁻² }`

T = `\frac {(100 m s⁻¹ ) (0.5)}{9.81 m s⁻² }`

T = 5.097 s

T = 5.1 s ----------------Ans.


Thus the time of flight of the bullet is 5.1 s


(c) Range of bullet = R = ?

The formula to find the Range of a projectile is given;

R = `\frac {vi² sin (2θ)}{g}`


R = `\frac {(50 m s⁻¹ )² sin (2ｘ30)}{9.81 m s⁻² }`


R = `\frac {(2500 m² s⁻² ) sin (60)}{9.81 m s⁻² }`


R = `\frac {(2500 m² s⁻² ) (0.866)}{9.81 m s⁻² }`


R = 220.699 m ----------------Ans.


Thus the Range of stone is 220.7 m




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