A projectile is shot on level ground with a horizontal velocity of 21 m/s and a vertical velocity of 39 m/s .
part1) Find the time the projectile is in the air. The acceleration due to gravity is 9.8 m/s2 .
Answer in units of s.
part 2)Find the range R. Answer in units of m.
part 3) Let the magnitude of v0 be the same as in part 1, but consider the angle θ required to obtain the maximum range.
Find the height attained for the maximum range.
Answer in units of m.
I got part 1 and 2, but I don't get 3????
I guess s = sqrt(21^2+39^2) = sqrt(441+1521) = 44.3 m/s
u = Vi = 44.3 /sqrt 2 = 31.3
cheat use energy
(1/2) m Vi^2 = m g h
h= Vi^2/(2g)
=31.3^2/19.6 = 50 meters
for max range initial speed s and angle A
u = horizontal speed =constant = s cos A
Vi = s sin A
v = Vi - g t
h = Vi t - .5 g t^2
h =0 at start and finish
0 = (Vi - .5gt) t
so finish t = Vi/(.5g)
range = u t= s cos A t
range = s cos A * Vi /(.5g )
= [1/(.5g)] s cos A s sin A = [s^2/(.5g)] cos A sin A
when is cos A sin A max?
cos A sin A = (1/2) sin 2A trig identity
max when 2A = 90 deg, A = 45 deg (could have guessed that I bet)
go back and fire it at 45 degrees up from horizontal
okay so for my final answer I got 64m? but I still think its wrong can you check for me?
ohhh I see thank you so much
In part 1 and 2, you already calculated the time of flight and the horizontal range of the projectile. Now, in part 3, you need to determine the height attained by the projectile when it achieves its maximum range. To do this, you'll need to analyze the projectile motion at different launch angles.
For a projectile launched at an angle θ with the same initial velocity as in the previous parts, the horizontal and vertical components of velocity can be calculated as follows:
Horizontal component (Vx): Vx = v0 * cos(θ)
Vertical component (Vy): Vy = v0 * sin(θ)
To find the time of flight (t) and maximum range (R) for a given launch angle, we can use the formulas:
Time of flight (t) = 2 * Vy / g
Maximum range (R) = Vx * t
Since we're interested in finding the angle θ that gives the maximum range, we can differentiate the maximum range equation with respect to θ and then solve for θ when the derivative is equal to zero:
dR/dθ = 0
d/dθ (Vx * t) = 0
Now, let's substitute the expressions for Vx and t back into the equation and differentiate:
d/dθ (v0 * cos(θ) * (2 * (v0 * sin(θ)) / g) ) = 0
Differentiating this equation requires the product rule and chain rule. Simplifying and solving for θ gives you the desired answer.
Please note that the mathematical derivation of the maximum range equation involves calculus, which might not be suitable for this platform. However, you can use the techniques mentioned to solve the equation and find the height attained for the maximum range.