Answer:
0.707m
Explanation:
from formula of range i.e R=Usin2Q/g
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:
101397.16 pa
Explanation:
The pressure recorded will be equal to pgh
Where p = density of mercury = 13.6x10^3 kg/m^ 3
g = acceleration due to gravity 9.81 m/s^2
h = height of mercury in the column = 760 mm = 760x10^-3 m
Pressure = 13.6x10^3 x 9.81 x 760x10^-3 = 101397.16 pa