please help!! projectile question!! (1 Viewer)

[HSC*is*near]

Velocity = 75 ms-1 (in the Vx direction. it is not the initial velocity. )
--------------------------------->>


height = 60m


1) Calculate the range of the projectile.
2) Calculate the vertical component of the velocity when the ball hits the ground
3) From the horizontal component and the vertical component calculate the velocity with which the projectile hits the ground.

(sorry cudnt put the diagram in this properly.. i tried but dats all i cud do... i hope it make's sense )

 

[zeropoint]

Velocity = 75 ms-1 (in the Vx direction. it is not the initial velocity. )
--------------------------------->>

/|\
| height = 60m
|
\|/



1) Calculate the range of the projectile.
2) Calculate the vertical component of the velocity when the ball hits the ground
3) From the horizontal component and the vertical component calculate the velocity with which the projectile hits the ground.

(sorry cudnt put the diagram in this properly.. i tried but dats all i cud do... i hope it make's sense )

Am I correct in assuming that the projectile is launched horizontally with speed 75 m/s from the edge of a 60 m high cliff?

If you want the range, you need to know how long the thing stays airborn, combine this with the fact that it's horizontal speed is unchaning allows you to calculate the range. From kinematics of constantly accelerated motion, you know that after a time t, a released object will fall a distance

h = 1/2 g t^2

in this case h = 60 m, so t = ( 2*h / g )^(1/2) = 3.5 s.

your range is going to be (75 m/s)(3.5 s) = 262 m.

2) In other words, you want to know the vertical speed after 3.5 s. It's just the derivative of the first equation,

Vy = -gt = -34 m/s.

3) Well, you know the components of the velocity vector at time of impact, they are Vx = 75 m/s and Vy = -34 m/s. So you're velocity in component notation is (75 m/s, -34 m/s). If you want the polar form, just draw the right-handed triangle and calculate the hypotenuse and the angle tan (theta) = 75/34

 

[Riviet]

2) Initially, u=0
So we use v²=u²+2as, where s is the displacement/distance.
v²=2as
Let's take a=9.8 in this question,
v²=2 x 9.8 x 60
=1176
v=34.29... (you should store all exact values in your calculator for later use and only round when you reach the final answer)
Therefore the ball hits the ground at 34ms^-1 vertically.

1) We use v=u+at and solve for t to find the time of flight:
When v=34.29...
t=(v-u)/a
=34.29/9.8 since u=0
=3.499...
Using the time of flight we can work out how far the ball has travelled horizontally in that time:
Distance = speed x time
= 75 x 3.499...
= 262.44...
Therefore horizontal range is 262m.

3) Draw a right triangle like the one below, label side AB as your horizontal component 75, side BC as your vertical component 34.
A_B
\aa|
.\a.|
..\a|
...\.|
aaC
Use Pythagoras' Theorem to find the hypotenuse, AC:
AC=sqrt(75²+34²)
=82.346...
Let @ be the angle between the horizontal and the hypotenuse AC that we just found. Find angle ACB using simple trigonometry and subtract it from 90^o to find the angle to the horizontal that the ball hits the ground.

Hope that helps.

 

[HSC*is*near]

thanks champions.... that helped heaps..... yay im happy... lol.. thanks