Answer:
S(t) = -4.9t^2 + Vot + 282.24
Step-by-step explanation:
Since the rocket is launched from the ground, So = 0 and S(t) = 0
Using s(t)=gt^2+v0t+s0 to get time t
Where g acceleration due to gravity = -4.9m/s^2. and
initial velocity = 39.2 m/a
0 = -4.9t2 + 39.2t
4.9t = 39.2
t = 8s
Substitute t in the model equation
S(t) = -49(8^2) + 3.92(8) + So
Let S(t) =0
0 = - 313.6 + 31.36 + So
So = 282.24m
The equation that can be used to model the height of the rocket after t seconds will be:
S(t) = -4.9t^2 + Vot + 282.24
<h3>Rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr</h3><h3><u>Solution:</u></h3>
Given that,
A motorboat travels 165 kilometers in 3 hours going upstream and 510 kilometers in 6 hours going downstream
Therefore,
Upstream distance = 165 km
Upstream time = 3 hours
<h3><u>Find upstream speed:</u></h3>

Thus upstream speed is 55 km per hour
Downstream distance = 510 km
Downstream time = 6 hours
<h3><u>Find downstream speed:</u></h3>

Thus downstream speed is 85 km per hour
<em><u>If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then</u></em>
Speed downstream = u + v km/hr
Speed upstream = u - v km/hr
Therefore,
u + v = 85 ----- eqn 1
u - v = 55 ----- eqn 2
Solve both
Add them
u + v + u - v = 85 + 55
2u = 140
u = 70
<em><u>Substitute u = 70 in eqn 1</u></em>
70 + v = 85
v = 85 - 70
v = 15
Thus rate of the boat in still water is 70 km/hr and rate of the current is 15 km/hr
Answer:
16
Step-by-step explanation:
8*2
Factor the numerator and factor the denominator.

Then you divide the numerator and the denominator by the common factor to reduce the fraction.