Question

22. A basketball is shot at a 70° angle with the horizontal with an initial velocity of 10 meters per second. Find the component form of the initial velocity.

A. v ~ (9.40,3.42)
B. v ~ (-3.42,9.40)
C. v ~ (3.42,-9.40)
D. v ~ (3.42, 9.40)
E. v ~ (-9.40,-3.42)
F. v ~ (9.40, -3.42)

Answer 1

component form requires the horizontal (x) and vertical (y) velocities

how do you think you can obtain those (hint: the relationship can come form the picture)

Answer 2

ok that didnt help me at all

Answer 3

......

Answer 4

i'm guessing this is for a trig class?
you should have gone over how to relate the sides of a triangle to the hypotenuse.
SOH CAH TOA ring a bell?

Answer 5

ya it rings a bell

Answer 6

so SOH CAH TOA is a memory aid which helps you remember how the sides of a right angle triangle are related through trig functions:
example, consider SOH:

\[\large \sin\theta = \frac{opposite}{hypotenuse}\]

meaning that if you have an angle (in your case, 70 degrees) than the ratio of the side opposite the angle and the hypotenuse is given by sin(70)

Answer 7

look at the equation, you'll see that there are 3 variables (opposite length, hypotenuse length, and the angle). If you are ever given two of the three you can solve the third by using this trig equation

Answer 8

from the drawing you can see that you are given the hypotenuse (10m/s) and the angle (70 degrees. solving this will give you the opposite side which, by looking at the drawing, is the same as the vertical component of velocity:

\[\large \sin(70^o)= \frac{opposite}{10}\]

try solving that and see what you get.

Answer 9

i think its D

Answer 10

are you guessing or did you solve the triangle using trig?

Answer 11

guessing, I dont understand but I gotta get this done tonight

Answer 12

what part don't you understand?