5.00kg
random = abcdefghijklmnopqrstuvwxyz
You are exerting 100N. Since there’s no NET force, then there must be exactly 100N pushing exactly back on your 100N to cancel it to exactly zero. Newton's first law states that whether a body is at rest or travelling in a straight line at a constant speed, it will remain at rest or continue to move in a straight line at a constant speed unless acted upon by a force.
We need to use the kinematic equation
S=ut+(1/2)at^2
where
S=displacement (+=up, in metres)
u=initial velocity (m/s)
t=time (seconds)
a=acceleration (+=up, in m/s^2)
Substitute values
S=displacement = 1.96-2.27 = -0.31 m (so that shot does not hit his head)
u=11.1
a=-9.81 (acceleration due to gravity)
-0.31=11.1t+(1/2)(-9.81)t^2
Rearrange and solve for t
-4.905t^2+11.1t-0.31=0
t=-0.02756 or t=2.291 seconds
Reject the negative root to give
t=2.29 seconds (to 3 significant figures)
Answer: d. 8.25 m/s
Explanation:
We are given that Current= 5 m/s in j direction
Velocity= 8 m/s i + 3 m/s j
Now, we have to find Jada's speed with respect to the water.
First we find Jada's velocity with respect to water
v= (8 i + 3 j) - (5 j)
v= 8i - 2 j
To find the speed, we take the magnitude of this velocity vector we have
|v|= 
|v|= 
= 8.246 m/s
which comes out to be around = 8.25 m/s
So option d is correct.
