A ball is launched into air. If the ball is launched with an initial height of 9 feet, what are the parametric equations for the ball's motion?

Question

the ball is launched with an initial velocity of 89 feet per second and an angle of 61∘ to the horizontal

select the correct answer below:

x(t)=−16t2+(89sin61∘)t+9, y(t)=(89cos61∘)t

x(t)=(89cos61∘)t, y(t)=−16t2+(89sin61∘)t+9

x(t)=−4.9t2+(89cos61∘)t+9, y(t)=(89sin61∘)t

x(t)=(89sin61∘)t, y(t)=−16t2+(89cos61∘)t+9

x(t)=(89cos61∘)t, y(t)=−4.9t2+(89sin61∘)t+9

Answer 1

Both x(t) and y(t) are of the general form

at²/2 + vt + s

where

a is the projection of the ball's constant acceleration on the axis,

v is the projection of the ball's initial velocity on the axis,

s is the projection of the ball's initial displacement on the axis.

In general, any vector with magnitude w at an angle α with the horizontal will have

projection w×cos(α) on the x-axis, and

projection w×sin(α) on the y-axis.

Now let's calculate the projections of the given quantities on the x-axis and y-axis.

Since the given quantities are in feet, we will use feet for all distances.

The ball is under gravitational acceleration of magnitude 32 ft/s².

The acceleration is vertical downward, forming an angle of 270° from the horizontal.

Its projection on the x-axis is 32×cos(270°) = 0.

Its projection on the y-axis is 32×sin(270°) = -32.

The ball has initial velocity 89 ft/s at angle 61° with the horizontal.

Its projection on the x-axis is 89×cos(61°).

Its projection on the y-axis is 89×sin(61°).

The ball has initial displacement of 9 ft an an angle 90° with the horizontal.

Its projection on the x-axis is 9×cos(90°) = 0.

Its projection on the y-axis is 9×sin(90°) = 9.

Therefore

x(t) = 89×cos(61°)×t

y(t) = -32×t²/2 + 89×sin(61°)×t + 9