Projectile Motion

Question

A cannon sends a projectile towards a target puttlitz (cjp3843) – Assign 8 2D and Projectile Motion – himes – (220222023) 3 a distance 1230 m away. The initial velocity makes an angle 38◦ with the horizontal. The target is hit. The acceleration of gravity is 9.8 m/s 2 . What is the magnitude of the initial velocity? Answer in units of m/s. Your answer must be within ± 3.0%

If the initial velocity is 111.4585585, How high is the highest point of the trajectory? Answer in units of m. Your answer must be within ± 3.0%

Answer 1

There are two motions taking place simultaneously: vertical and horizontal.

Horizontal is very simple: constant speed motion. Horizontal distance:

L = v_{i *} cos38^o _* t

where v_i is initial velocity, and t is time of flight.

Vertical motion is more complex: first half of flight it is decelerated with g, second half it is accelerated with g.

i.e. vertical velocity is not constant, it changes with time, i.e. it is a function of time

v(t) = v_{i *} sin38^o - g _* t

The key is to realize that at top of the trajectory vertical speed is 0. This moment of time t₁ is when first half of the flight is over and second half of flight begins. If we plug in this specific moment of time t₁ in the equation above, we get

v(t₁) = v_{i *} sin38^o - g _* t₁ = 0

From this we can compute t₁:

t₁ = v_{i *} sin38^o / g

Duration of the 2nd half is exactly equal to the duration of the 1st half of flight. Therefore total time of flight t is:

t = 2 * t₁ = 2 _* v_{i *} sin38^o / g

We can plug it in the very first equation above, and know that L = 1230m.

L = v_{i *} cos38^o _* t;

L = v_{i *} cos38^o _* 2 _* v_{i *} sin38^o / g

L = 2 _* v_i² _* cos38^o _* sin38^o / g

v_i = [ L * g / (2_* cos38^o _* sin38^o ) ] ^1/2

v_i = 35.2m/s

Problem contains the following statement: "Your answer must be within ± 3.0% ". So what is the ask?

3% of 35m/s is 1m/s

It probably implies that answer should be rounded to whole m/s, i.e. v_i = 35m/s. But 35.2m/s is better than 3%. Would you be penalized by giving 35.2m/s as an answer? I don't know... this is a perfect question for the professor's office hour. She would love to see you come in and ask this question. Do it.

Part 2. The highest point of the trajectory. This is a question about the vertical motion, which is accelerated motion, remember? Displacement during accelerated motion, h

h(t) = v_{i *} sin38^o_* t - g _* t² / 2

If we plug in t₁, which is time when projectile is at the top of the trajectory, then we compute maximum height

h_max = ( v_{i *} sin38^o )² / g - (g / 2) (v_{i *} sin38^o / g )²

h_max = 0.5 _* ( v_{i *} sin38^o )² / g = 23.96m

Computing 3% of it: 0.03 * 23.96 = 0.7m

Therefore I am rounding 23.96m to 24m. But again, please verify with your teacher that this is what he is implied by "+3%" statement