Question

A ball is kicked at a 50 degree angle with an initial velocity of 21 ft/s.
Write the equation that models the height.
What is the maximum height reached by the object?

Answer 1

\[Hmax = \frac{ u^2 \sin^2 \theta }{ 2g }\]

Answer 2

Using S = ut + at^2/2

Height = usin theta * t - g t^2/2

Answer 3

im supposed to use this equation h(t) =-16t^2+v Sin (50)+h

Answer 4

did you calculate the y component of the initial velocity

Answer 5

equation:h(t)= -16t^2+16.08t+0
For the max height i got 101.8 ft

Answer 6

is it starting from initial height of 0 ?

Answer 7

yes

Answer 8

Would that be correct?

Answer 9

not sure, doesnt look right. how did you get that?

Answer 10

\[h = 16.08t - 16t^{2}\]

the max height occurs at time t = 0.5025 and the max height is onlly about 4.04 ft

Answer 11

i used -b/2a wich would be -16/-32=0.5 and plug that back in the equation

Answer 12

-b/2a gives -16.08/(2(-16)) = 0.5025

Answer 13

plugging that in gives a max height of 4.04 ft ?

Answer 14

oh i see where i made my mistake.

Answer 15

Im supposed make up a problem what should i change to make it sound like a more realistic height?

Answer 16

i mean a higher height

Answer 17

change the angle to something steeper, and add more initial velocity

Answer 18

how high are you trying to kick it ?

Answer 19

Something like 9 9ft

Answer 20

9

Answer 21

Thank you very much sir. :)