Grade 12 Physics: Projectile Question?

Question

I'm not sure how to solve for the velocity of a projectile when the initial and final points are not on the same horizontal level. Any general tips, or an explanation of the answer to the following question would be appreciated. Thanks.

Fizzicks has climbed to a ledge 270 metres above the base of a mountain. Hamilton, standing on level ground, uses his slingshot to send a canteen to Fizzicks, a horizontal distance of 390 metres away. The canteen is launched at an angle of 70 degrees to the horizontal. What should be the launch speed of the canteen?

Answer 1

Well you know that the the vertical component of the motion is sin(70)V and the horizontal component is cos(70)V where V is the initial velocity.

You are looking that

sin(70)V*t - 4.9t^2 = 270

and

cos(70)V*t = 390

The easiest way to solve this is to partially substitute t. Solve for t in the second equation to arrive at t = 390/(cos(70)V).
Partially substituting this yields:

tan(70)390 - 4.9t^2 = 270

1100 - 4.9t^2 = 270

t^2 = 170
t = 13 s

Knowing the time you can find the velocity using the horizontal component equation.

V = 390/(cos(70)*13 s) = 88 m/s.

Edit: Whoa, I didn't realize it took my so long to type up the answer...

Answer 2

In two-dimensional kinematics problems, each dimension is an independent problem except for time, which is shared by both. So a common approach is to write down the equations of motion in each dimension, use one to find the value for (or at least and expresion for ) the elapsed time, and then carry that time into the other dimension.

The vertical motion is

( y - yo ) = voy t - 1/2 g t^2

y - yo is the vertical displacement (in this case, -270 m )
voy is the initial vertical velocity ( vo sin 70 )
g is the acceleration of gravity, 9.8 m/s^2
t is the elapsed time

The horizontal motion is

x - xo = vox t

x - xo is the horizontal distance covered, 390 m
vox is the initial horizontal velocity, vo cos 70
t is, again, the elapsed time

Solve the second equation for time...

t = ( x - xo ) / ( vo cos 70 )

...and substitute this expression for t in the first equation. Now the only unknown is the desired launch speed v.

Answer 3

attitude 30 tiers Sin 30 = 0.5 Horizontal velocity = 0.5* sixteen.4 Vertical velocity = sixteen.4 - Horiz vel. Time of flight is time to attain max hight and return to floor Hor displacement = 8.2*t