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A cannon fires a 0.652 kg shell with initial

velocity vi = 12 m/s in the direction = 61 ◦

above the horizontal.

The shell’s trajectory curves downward be-

cause of gravity, so at the time t = 0.473 s

the shell is below the straight line by some

vertical distance h.

Find this distance h in the absence of

air resistance. The acceleration of gravity is

9.8 m/s2 .

Answer in units of m

Verified Answer :

syed514

Mass have no effect for the projectile motion and  u want to know the  height "h"   
first,
        find the vertical and horizontal components of velocity 
 vertical component of velocity = 12 sin 61                  
horizontal component of velocity = 12 cos 61
now for the vertical motion ;               
             S = ut + (1/2) at^2
where
 s = h 
u = initial vertical component of velocity 
t = 0.473 s 
a = gravitational deceleration (-g) = -9.8 m/s^2    
       
         h=[12×sin 610×0.473]+[−9.8×(0.473)2] 

u can simplify this and u will get the answer

h=.5Gt2 

H=1.09m

Answer:

height= 6.429m

Explanation:

h is equall to Max height traveled by trajectory body.

Max height= (U²Sin²tita)/2g

Max height = (12²(sin61)²)/(29.8)

Max height =( 1440.875)/19.6

Max height= 6.429 m

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