Answer:
vo=5.87m/s
Explanation:
Hello! In this problem we have a uniformly varied rectilinear movement.
Taking into account the data:
α =69.2
vf = 10m / s
h=2.7m
g=9.8m/s2
We know we want to know the speed on the y axis.
We calculate vfy
vfy = 10m / s * (sen69.2) = 9.35m / s
We can use the following equation.

We clear the vo (initial speed)


vo=5.87m/s
Answer:
The smallest distance the student that the student could be possibly be from the starting point is 6.5 meters.
Explanation:
For 2 quantities A and B represented as
and 
The sum is represented as
For the the values given to us the sum is calculated as

Now the since the uncertainity inthe sum is 
The closest possible distance at which the student can be is obtained by taking the negative sign in the uncertainity
Thus closest distance equals
meters
Answer:
The correct option is C
Explanation:
According to third equation of motion, v
2
=u
2
+2ax
Here, u=0 m/s
a=−g and x=−h
Negative sign indicates downward direction. Displacement and acceleration both are downwards.
So,v=±
2(−g)(−h)
We take minus sign because it is downwards.
v=−
2gh
After bouncing. velocity becomes 80% of v, i.e.,
v
′
=+0.8
2gh
(positive sign because the direction of ball has reversed after bouncing and is upwards.
Applying third equation of motion again, for u=v
′
, v=0 and a=−g
v
2
=u
2
+2×a×x
Thus,
0=0.64(2gh)+2(−g)x
or
x=0.64h
Answer:

Explanation:
Given data
Electromagnetic waves from the sun is I=1.4kW/m² at 80%
Area a=(0.30×0.51)m²
Time t=1.30 hr
To find
Energy E
Solution
The energy received by your back is calculated as:
