projectiles anyone??

Question

A projectile is launched at an angle of 35° above the horizontal with a speed 107 m/s. At the highest point in its trajectory, the vertical displacement above the ground is?

Answer 1

Given:
theta = 35 degrees
vi = 107 m/s

Find:
dy = ?

(I'M SO SORRY BUT I DON'T HAVE A CALCULATOR WITH ME!)

Solution:

viy = vi (sin theta)
viy = 107 m/s ( sin 35 degrees)
viy = ?

Then,
t = vi (sin theta) / g

So,
dy = viy(t) +1/2gt^2

Good luck and God bless!

Answer 2

This is a kinematics problem in two dimensions. However, you are not asked for anything in the horizontal dimension, so all we care about is the vertical dimension. The problem begins when the projectile is fired and ends when the projectile reaches the peak.

The vertical motion is:

v = vo - g t
y = yo + vo t - 1/2 g t^2

where

v is the final velocity (zero)
vo is the initial velocity ( 107 sin 35 degrees)
g is the local acceleration of gravity ( 9.8 m/s^2)
t is the elapsed time (unknown)
y is the final height (what you're looking for)
yo is the initial height (zero)

In the first equation, you know everything except time. Solve for time, then use that value in the second equation to solve for height.

Answer 3

An object traveling for 1 second at 107 m/s would travel 385 kph, 21063 ft per min, 351ft per second, 7021 yards per min, 239mph, 208 knots...... so if the Horizontal plane is 0 degrees it should travel for a distance of 351 ft. Altitude is based on the height of the horizontal on a triangular vecter. If you were to plot a graph you will find the Height. You can draw that with a protractor.

Answer 4

firstly, resolve the 107 m/s speed into horizontal and vertical components using trigonometry and a vector diagram.
as you will find;
let horizontal speed be Vx,
let vertical speed be Uy
Ux = 107cos35
Uy = 107sin35
these represent initial speeds
since at highest point, the vertical speed must reach zero, before turning negative (because its direction is now opposite because its falling)

so with initial vertical speed being Uy, and final vertical speed being 0, we also have vertical acceleration which is gravity (acting down, so a = g = -9.81 m /s)
so with u, v, and a, we still need s (displacement)

so considering motion only in the vertical direction, we use
v^2 = u^2 + 2as, (where a = -g)
rearrange in terms of s, and substitute for values,

we could have used time of flight, but i think this is how i would solve it to avoid extra steps.