How Fast and How Long Does a Pebble Travel in a Projectile Experiment?

[carltonblues]

God I hate projectiles..

A PHYS1001 student uses a slingshot to project a pebble at her lecturer
(shoulder height) who is standing 40 m away. After extensive experiments she
finds that to hit the target she must aim 4.85 m above the target. Determine the
velocity of the pebble on leaving the slingshot and the time of flight.

Can some one start me off, with the correct formula to use. I am stumped..

 

[xanthym]

God I hate projectiles..

A PHYS1001 student uses a slingshot to project a pebble at her lecturer
(shoulder height) who is standing 40 m away. After extensive experiments she
finds that to hit the target she must aim 4.85 m above the target. Determine the
velocity of the pebble on leaving the slingshot and the time of flight.

Can some one start me off, with the correct formula to use. I am stumped..

The projectile is required to hit the target at the same height it leaves the student. Furthermore, the Angle θ (above horizontal) at which the projectile leaves the student is given by:
tan(θ) = (4.85)/(40) = (0.121) ::: ⇒ θ = (6.91 deg)

The standard equations for projectile height z(t) under constant gravitational acceleration "g" and for horizontal distance "x" are given by:
z(t) = z₀ + v_z0t - (1/2)gt² :::: Eq #1
x(t) = x₀ + v_x0t :::: Eq #2

Let T be the time of flight. The problem requires that {z(T) = z₀} and that {x(T) - x₀ = (40 m)}. Placing these into Eqs #1 & #2 we get:
z(T) = z₀ = z₀ + v_z0T - (1/2)(9.81)T²
⇒ 0 = v_z0T - (1/2)(9.81)T²
⇒ 0 = v_z0 - (4.91)T
⇒ T = v_z0/(4.91) :::: Eq #3

x(T) - x₀ = 40 = v_x0T
⇒ T = (40)/v_x0 :::: Eq #4

We now multiply Eq #3 by Eq #4:
T² = {40/4.91}*{v_z0/v_x0} =
= (8.15)*tan(θ) =
= (8.15)*(0.121) =
= (0.986)
⇒ T = (0.993 sec)

Placing this last result into Eq #4 and solving for v_x0:
(0.993) = (40)/v_x0 ::: ⇒ v_x0 = (40)/(0.993) = (40.3 m/sec)
⇒ v₀ = v_x0/cos(θ) = (40.3)/cos(6.91 deg)
⇒ v₀ = (40.6 m/sec)


~~

 

[carltonblues]

life saver!

 

[maverick280857]

There's another way, though not too different from xanthym's.

The trajectory that the pebble follows is a parabola under the (assumed constant) gravitational field of earth. If the pebble were projected with an initial velocity \vec{u} at an initial elevation \alpha (computed just as xanthym did) then the equation of the trajectory turns out to be,

y = xtan\alpha - \frac{1}{2}\frac{gx^2}{u^2}sec^2\alpha

In principle you can find u directly from this equation given y = 0, x (= range) and alpha. Once you have that, use Range = u\cos\alpha*Time of Flight to find the time of flight.

The equation here isn't really a shortcut as in some cases, the algebraic calculations are easier if you follow the step-by-step fundamental method. Yet its worthwhile to know the equation, as it relates all the parameters of motion and gives you y as a function of x.

Cheers
Vivek

PS--It would be good exercise for you to derive this equation of trajectory yourself. Hint: xanthym's equations are sufficient to do this.

 

[Am kuls]

look at http://www.goalfinder.com/downloads.asp for a simulation of projectile, great for learning and fun to work with