Answer: a) 7.56m/s b) 6.245m/s on its way up and -6.245m/s on its way down.
Explanation:
a) Velocity after 2 seconds
H = 15t - 1.86t^2
V(0) = 15 m/s
Now let’s first differentiate h with respect to t , we have
d/dt x h = d/dt (15t - 1.86t^2)
= 15 d/dt x t - 1.86 d/dt x t^2
= 15 x 1 - 1.86(2t) = 15 - 3.72t
t = 2 sec
Therefore
V(t) = 15 - 3.72(2) = 7.56 m/s
b) if height is 25m
H = 15t - 1.86t^2
Therefore h = 25, we have
25 = 15t - 1.86t^2
15t - 1.86t^2 - 25 = 0
1500t - 186t^2 - 2500 = 0
-186t^2 + 1500t - 2500 = 0
Using quadratic equation to find the roots of equation
First root t1 = 2.3535
Second root T2 = 5.71102
Now, at t1,
V(t1) = 15 - 3.72 (2.3535) = 6.245 m/s
The velocity of the rock on its way up is 6.245 m/s
V(T2) = 15 - 3.72 (5.71102) = -6.245 m/s
The velocity of the rock on its way down is -6.245 m/s
The velocity of the rock after 2seconds is 7.56m/s
The velocity of the rock when its height is 25 m on its way up is 6.072m/s
Given the height reached by the rock modeled by the equation;
[tex]H = 15t - 1.86t^2[/tex]
t is the time taken
The velocity of the rock is gotten by differentiating the function with respect to t as shown:
[tex]\frac{dh}{dt}=15- 3.72t\\v(t)=15- 3.72t\\[/tex]
a) To get the velocity of the rock after 2 secs, we will substitute t = 2 into the velocity function as shown:
[tex]v(t)=15-3.72t\\v(2)=15-3.72(2)\\v(2)=15-7.44\\v(2)=7.56m/s[/tex]
Hence the velocity of the rock after 2seconds is 7.56m/s
b) Recall that [tex]h=15t-1.86t^2[/tex]
If the rock is 25m on its way up, then;
[tex]25 = 15t-1.86t^2\\-1.86t^2+15t - 25 =0\\1.86t^2-15t+25 =0[/tex]
On factorizing, t = 2.4s and 5.7s
Substitute the times into the velocity function to get the required velocity:
[tex]v(t_1) = 15-3.72t_1\\v(2.4) = 15-3.72(2.4)\\v(2.4) = 15-8.928\\v(2.4) = 6.072m/s\\[/tex]
At the time t = 5.7s
[tex]v(t_2) = 15-3.72t_2\\v(5.7) = 15-3.72(5.7)\\v(5.7) = 15-21.204\\v(5.7) = -6.204m/s\\[/tex]
Hence the velocity of the rock when its height is 25 m on its way up is 6.072m/s
Learn more here: https://brainly.com/question/1376074?section=related_q
