Question

Does the acceleration vector change at all throughout the projectile’s flight? Why or why not?

Compare two complementary angles such as 30? and 60?. What is the same? What is different?

How does changing the initial velocity affect the projectile’s motion?

How does the time spent in the air depend on the launch angle?

Answer 1

During a projectile motion, the acceleration acting on the object is acceleration due to gravity g and this acceleration will always act towards the center of earth of magnitude g = 9.81 m/s² i.e. Vertically downward throughout the projectile motion. so, Acceleration vector does not change at all throughout the projectile's flight.

Comparing two complementary angles as angle of launch of projectile as 1 =  30? and 2 = 60? :-

for both the angles , vertical acceleration as acceleration due to gravity as g acting in vertically downward direction will remain same .

As we kmow,

Horizontal Range, R =  u²* sin 2 /g

for   1 =  30?, R1 =  u²* sin (2*30) /g = 0.866 *u² / g

for   2 = 60?, R2 =  u²* sin (2*60) /g = 0.866 *u² / g

só, for two given value of angles, horizontal range will also remain the same

As we know maximum height achieved in a projectile, H_max = u²* sin²/2g

for   1 =  30?, H1 =  u²* sin² (30) /g = 0.25 *u² / 2g

for   2 = 30?, R1 =  u²* sin² (60) /g = 0.75 *u² / 2g

só, maximum height achieved by the object will be different for two given angles,

Time of flight, T =  2u* sin/g

for   1 =  30?, T1 =  2u* sin30 /g = u/g

for   2 = 60?, T2 =  2u* sin60 /g = 1.732 u/g

so, for object will remain in the projectile for different time for two given angles.

from above equation we can see that maximum Height , H u²,

so we can say that by increase the magnitude of initial velocity, maximum height will increased and vice-versa.

Horizontal Range , R u²

so we can say that by increase the magnityde of initial velocity, horizontal range will increased and vice-versa.

Time of flight, T u,

so we can say that by increase the magnitude of initial velocity, the object will remain in the air for greater time and by decreasing initial velocity, the object will remain in the air for lesser time.

According to equation of motion, a s

H = u_yt + 1/2 * a_y*t²

the object will reman in the air till the object will return back to the surface i.e. H = 0,

u_y = u * sin

and a_y = -g

so, 0 =  u * sin*t - 1/2*g*t²

so, t² = 2t* u * sin/g

hence time of flight, t = 2 u * sin/g

so, here, t    sin

so, by increasing value of , time spent in the air will increased and by decreasing value of , time spent in the air will decreased.