how can this be done?

Question

A ball is thrown upward from a platform 5.2 m high with a speed of 15 m/s at an angle of 40° from the horizontal. What is the magnitude of its velocity when it hits the ground? Note: You need to find the time taken first by solving a quadratic equation. Answer is t = 2.4 s.

Answer 1

1st step is to break velocity into horizontal and vertical components (think right triangle here):

vertical velocity = 15 * sin(40) = 9.642 m/s
horizontal velocity = 15 * cos(40) = 11.491 m/s

You could solve a quadratic equation to find the answer. My method has easier math. Just break the flight of the ball into the up and down components.

Part 1:
Figure out how many seconds before the rock's vertical velocity gets to zero (peak of arc):
time up = 9.642 / 9.8 = 0.984 sec

Find the average vertical velocity during that time with linear acceleration from 9.642 m/s to zero:
Vavg up = 9.642 / 2 = 4.821 m/s

Find the distance based on average speed and time:
distance up = 4.821 * 0.984 = 4.743 meters above the platform


Part 2:
So we know that the ball is at 9.943 meters above the ground with zero vertical velocity. Now, figure out what it's vertical velocity will be when it hits the ground.

d = distance travelled (-9.943)
Vgnd = vertical velocity when it hits
Vavg = average velocity on the way down
t = time from peak to hitting ground
a = gravitational acceleration (-9.8m/s^2)

We know:
Vavg = d/t
Vgnd = 2 * Vavg
Vgnd = a * t

2 * -9.943/t = -9.8t
-19.886 = -9.8t^2
t^2 = 2.029
t = 1.425 seconds

Vgnd = -9.8 * 1.425 = -13.96 m/s


Now convert back to a total velocity using the original horizontal component which isn't changed (right triangle again):

Vtotal = sqrt(11.491^2 + 13.960^2) = 18.08 m/s
Angle = inverse tan(13.960/11.491) = 50.5 deg (below the horizontal)

Answer 2

theoretically on the way down when it crosses the x axis
at + 5.2m it will be travelling at 15 m/s at an angle of 40 degrees from the horizontal.
acceleration from there continues the rest of the way down with a vertical component of 5.2 m x gravity added to trig function w/e

Answer 3

This has to do with sine (opposite/hypoteneuse), cosine (adjacent/hypoteneuse) and tangent (opposite/adjacent). There are tables in the back of your math book for this. Or check out the link below.

Answer 4

nuts

Answer 5

what