Question

i have the first part but i am confused on the b and c portion of this question

Answer 1

We are given that an object follows the next parametric equations:

$\begin{gathered} x=vcos\theta t \\ y=vsin\theta t+k-16t^2 \end{gathered}$

Where:

$\begin{gathered} x=\text{ horizontal distance} \\ y=\text{ height of the object} \\ v=\text{ initial velocity} \\ \theta=\text{ initial angle} \\ t=\text{ time} \end{gathered}$

Now, we substitute the given values for the horizontal reach we get:

$x=\lparen128cos60)t$

For part B we are asked to determine the time that the object would take to reach a horizontal distance of 400 feet. To do that we will substitute the value of "x = 400" in the equation:

$400=\operatorname{\lparen}128cos60)t$

Now, we solve for the time "t" by dividing both sides by "128cos60":

$\frac{400}{128cos60}=t$

Solving the operations:

$6.25=t$

Therefore, it would take 6.25 seconds for the object to reach the 400 ft.

Part C we are asked to determine the height when the object reaches the 400 ft. To do that we use the equation for the height "y":

$y=vs\imaginaryI n\theta t+k-16t^2$

Substituting the values we get:

$y=8+128sin60t-16t^2$

Now, we substitute the value of time "t = 6.25s" which is the time it takes the object to reach the 400 ft:

$y=8+128s\imaginaryI n60\left(6.25\right)-16\left(6.25\right)^2$

Solving the operations:

$y=75.82$

Therefore, the height is 75.82 ft.