Question

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????

Answer 1

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

Answer 2

okay so for my final answer I got 64m? but I still think its wrong can you check for me?

Answer 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

Answer 4

ohhh I see thank you so much