Question

Consider the motion of a projectile. It is
fired at t = 0. Its initial speed is 11 m/s
and its initial projection angle is 50◦ from the horizontal. How do I find the max height?

Answer 1

11 sin(50) is your initial up velocity

Answer 2

and since its in meters id go with g=4.9

Answer 3

My homework tells me that g=9.8ms^2 why would you half that?

Answer 4

-g/2 t^2 +Vi t + Hi where the highest point happens at:

-Vi/-g

Answer 5

its a math thing; they give you acceleration and you have to integrate it up to position

Answer 6

Oh okay, I'll trust you on that bit. So, now what numbers go where? I found V, so, am I solving for time/ Or the variable Hi?

Answer 7

a(t) = some constant:

int a(t) = v(t) = c1 t + c2

int v(t) = h(t) = c*t^2/2 + c2 t + c3

Answer 8

the Hi is initial height; lets assume 0 unless they give you it otherwise

Answer 9

Its 0.

Answer 10

Vi is the speed times the up vector of an angle of 50 degrees; up is sin

11 sin(50) t then

Answer 11

and gravity ups to -, to get the thing back down to earth again, 9.8/2 t^2 to put it all toghter

height(t) = -4.9 t^2 +11 sin(50)

Answer 12

Why are we solving for initial velocity if the problem gave it to me as 11m/s?

Answer 13

it didnt give us initial velocity perse; it gaves us the initial speed of the object going in a certain direction; at a 50 degree angle to the ground; that angle has 2 parts; out and up

Answer 14

to find out how far up it goes, we need to isolate the up portion

Answer 15

up velocity = 11 sin(50) and you need to find out if that needs to be in radians or degrees