Question

How far does an object travel horizontally at 35ft/s use the equation
x(t)=vcos(45)t

Answer 1

x(t)=vcos(35)t

Answer 2

I think they're saying \[
v =35 \\
x(t) = 35\cos(45^\circ )t
\]

Answer 3

Yes i cant find a way to do it! Please help

Answer 4

You need to figure out \(t\). I assume there's gravity involved.

Answer 5

Yes

Answer 6

\[
y(t) = \frac{1}{2}at^2+v_{y0}t+y_0
\]

Answer 7

\[
y_0 = 0 \\
v_{y0} = \sin(45^\circ) \\
a = -9.8
\]

Answer 8

It's a quadratic equation. Find the roots. One of them is obviously \(0\)

Answer 9

\[R = \frac{ u^2 \sin2\theta }{ g }\]

u = 35

theta = 45

2theta = 90

Answer 10

\[R = \frac{ u^2 *1}{ g }\]

Answer 11

Whoops, my \(a\) was in metric... you should use \(a=32\)

Answer 12

Oh ok that confused me

Answer 13

\[
y(t)= \left[ \frac{1}{2}(32)t-35\sin(45^\circ) \right]t
\]

Answer 14

@El_Pollo Can you solve for \(t\)?

Answer 15

Set \(y(t) = 0\)

Answer 16

t=1.54

Answer 17

Now find \(x(1.54)\) given the equation above.