Question

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

Answer 1

height= 6.429m

Step-by-step explanation:

h is equall to Max height traveled by trajectory body.

Max height= (U²Sin²tita)/2g

Max height = (12²(sin61)²)/(2*9.8)

Max height =( 144*0.875)/19.6

Max height= 6.429 m

Answer 2

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