Question

While trying out for the position of pitcher on your high school baseball team, you throw a fastball at 84.9 mi/h toward home plate, which is 18.4 m away. How far does the ball drop due to effects of gravity by the time it reaches home plate? (Ignore any effects due to air resistance and assume you throw the ball horizontally.)
==>_______m

Answer 1

@Jemurray3

Answer 2

The first thing you need to know is how long the ball was in the air. To do that, you should divide the distance it travels by its horizontal velocity. However, the velocity and position are in incompatible units -- first convert the velocity to m/s.

Answer 3

so the conversion is 37.95 m/s

Answer 4

To find the time it takes to go from the pitcher's mound to the plate, divide the distance travelled by the velocity.

Answer 5

so 18.4/37.95?

Answer 6

@Jemurray3

Answer 7

Yes. Now that you have the time, you know that the distance it falls in the y direction is given by the equation
\[ \Delta y = -\frac{1}{2} gt^2 \]
where g is the gravitational constant 9.81 m/s^2.

Answer 8

so it would equal 2.4?

Answer 9

No. Check your work.

Answer 10

i forgot the negative.... so -2.4?

Answer 11

Nope.

Answer 12

change in y= -1/2*9.81*0.49^2 = -1.18?

Answer 13

@Jemurray3

Answer 14

yes.

Answer 15

that was apparently wrong though ;/

Answer 16

if it's asking how far the ball drops, perhaps it would like you to disregard the minus sign.

Answer 17

so a positive 1.18?

Answer 18

mhmm

Answer 19

still wrong...

Answer 20

Using the proper number of significant figures gets me 1.15 meters.

Answer 21

that was correct. did I calculate wrong?

Answer 22

don't round your answers in the middle of the calculation.

Answer 23

oh i seee. okay.. Thank you!!!(: