To determine
To find: The velocity of the rock after 2 s if a rock is thrown with the velocity of 15 m/s and the height of the rock after t seconds is h(t)=15t−1.86t2.
Answer
The velocity after 2 s is 7.56 m/s.
Explanation
The height of the rock is h(t)=15t−1.86t2.
Find the derivative of h(t) and obtain the velocity.
v(t)=h'(t)=15−3.72t
Thus, the velocity is v(t)=15−3.72t (1)
Substitute t = 2 in equation (1) and obtain the velocity after 2 s.
v(2)=15−3.72(2)=15−7.44=7.56
Thus, the velocity after 2 s is 7.56 m/s.
To determine
To find: The velocities of the rock when it reaches 25 m on its way up and on its way down.
Answer
The velocity of the rock when its height is 25 m on its way up is 6.24 m/s.
The velocity of the rock when its height is 25 m on its way down is −6.24 m/s.
Explanation
Given:
The rock is thrown with the velocity of 15 m/s and the height of the rock after t seconds is h(t)=15t−1.86t2.
Calculation:
Set h(t)=25 and obtain the value of t.
15t−1.86t2=251.86t2−15t+25=0
Use quadratic formula and simplify further.
t=15±152−4(1.86)(25)2(1.86)=15±225−1863.72=15±493.72=15±73.72
Therefore, the solutions are t≈2.35 and t≈5.71.
Thus, the ball reaches 25 ft on its way up at t = 2.35 s and on its way down at t = 5.71 s.
From part (1), v(t)=15−3.72t.
Substitute t = 2.35 for v(t) and obtain the velocity on its way up.
v(2.35)=15−3.72(2.35)=15−8.742≈6.24
Thus, the velocity of the rock when it reaches 25 m on its way up is 6.24 m/s.
Substitute t = 5.71 for v(t) and obtain the velocity on its way up.
v(5.71)=15−3.72(5.71)=15−21.2412≈−6.24
Thus, the velocity of the rock when it reaches 25 m on its way down is −6.24 m/s.