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
The answer is going to be B
Answer:
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<h2><u><em>
D</em></u></h2>
Explanation:
3) For a displacement time graph, straight line denotes constant speed. For a velocity time graph, the graph parallel to time axis denotes constant speed. Hence, the correct option is a) and d).
Answer:
20n
Explanation:
60n - 40n = efficiency of pulley system
20n = efficiency of pulley system
