If a rock is thrown vertically upward from the surface of Mars with velocity of [tex]$25 m / s$[/tex], its height (in meters) after [tex]$t$[/tex] seconds is [tex]$h=25 t-1.86 t^2$[/tex].
(a) What is the velocity (in [tex]$m / s$[/tex]) of the rock after 2 s?
(b) What is the velocity (in [tex]$m / s$[/tex]) of the rock when its height is 70 m on its way up? On its way down? (Round your answers to two decimal places.)
Answer (1)
Find the velocity function by taking the derivative of the height function: v ( t ) = 25 − 3.72 t .
Calculate the velocity at t = 2 seconds: v ( 2 ) = 17.56 m / s .
Solve for t when h ( t ) = 70 meters, obtaining t 1 ≈ 3.976 (up) and t 2 ≈ 9.464 (down).
Calculate the velocities at these times: v ( 3.976 ) ≈ 10.21 m / s (up) and v ( 9.464 ) ≈ − 10.21 m / s (down). The velocity on the way up is 10.21 m / s and on the way down is − 10.21 m / s .
Explanation
Finding the Velocity Function The height of the rock is given by the equation h ( t ) = 25 t − 1.86 t 2 . To find the velocity, we need to find the derivative of the height function with respect to time, v ( t ) = h ′ ( t ) .
Calculating the Derivative Taking the derivative of h ( t ) with respect to t , we get:
v ( t ) = d t d ( 25 t − 1.86 t 2 ) = 25 − 2 ( 1.86 ) t = 25 − 3.72 t
Velocity at t=2 seconds (a) To find the velocity of the rock after 2 seconds, we substitute t = 2 into the velocity function:
v ( 2 ) = 25 − 3.72 ( 2 ) = 25 − 7.44 = 17.56
Setting up the Quadratic Equation (b) To find the velocity when the height is 70 m, we first need to find the time(s) when the height is 70 m. We set h ( t ) = 70 :
25 t − 1.86 t 2 = 70
Rearranging the equation, we get a quadratic equation:
− 1.86 t 2 + 25 t − 70 = 0
Solving for Time We can solve this quadratic equation using the quadratic formula:
t = 2 a − b ± b 2 − 4 a c
where a = − 1.86 , b = 25 , and c = − 70 .
t = 2 ( − 1.86 ) − 25 ± 2 5 2 − 4 ( − 1.86 ) ( − 70 ) = − 3.72 − 25 ± 625 − 520.8 = − 3.72 − 25 ± 104.2
t = − 3.72 − 25 ± 10.2078
Calculating the Two Times We have two possible values for t :
t 1 = − 3.72 − 25 + 10.2078 = − 3.72 − 14.7922 ≈ 3.976 seconds (on the way up)
t 2 = − 3.72 − 25 − 10.2078 = − 3.72 − 35.2078 ≈ 9.464 seconds (on the way down)
Calculating the Velocities Now we find the velocity at these times:
v ( t 1 ) = v ( 3.976 ) = 25 − 3.72 ( 3.976 ) = 25 − 14.791 ≈ 10.21 m / s (on the way up)
v ( t 2 ) = v ( 9.464 ) = 25 − 3.72 ( 9.464 ) = 25 − 35.206 ≈ − 10.21 m / s (on the way down)
Final Answer Therefore, the velocity of the rock after 2 seconds is 17.56 m / s . When the height is 70 m, the velocity on the way up is approximately 10.21 m / s , and the velocity on the way down is approximately − 10.21 m / s .
Examples
Understanding projectile motion is crucial in fields like sports science and engineering. For instance, when designing a catapult or analyzing a baseball pitch, engineers and scientists use equations of motion to predict the trajectory, velocity, and impact point of projectiles. This problem demonstrates how to calculate the velocity of a projectile at a specific time and height, which is essential for optimizing performance and ensuring safety in various applications.
