The ideas required to deal with the given problem are kinematic equations the motion.

You are watching: Which projectile spends more time in the air, the one fired from 30∘ or the one fired from 60∘?

Initially, achieve the expression for time of trip for a projectile released with an edge by using kinematic equation that relates early speed, final speed, acceleration and also time. Then, usage the obtained expression because that time of trip to identify the projection angle with more time of flight.

The kinematic equations space a collection of equations the are offered to describe the motion of an object moving with constant acceleration.

The kinematic equation the relates final velocity v, early stage velocity , acceleration a and also time is offered by complying with expression.


Assume a particle launched at angle

with a velocity
. The vertical component that velocity of the bit is given by adhering to expression.

The horizontal ingredient of velocity

of the particle is provided by complying with expression.


Assume a projectile start as presented in the listed below figure.


is the initial velocity, v is the velocity in ~ maximum height and also is the angle of projection.

Use the following kinematic equation in upright direction to find the time to reach maximum height.



is the upright component of early velocity, is the upright acceleration, and also is the moment to reach maximum height.

Substitute 0 for v,

because that
for in the over equation to settle for time.


The total time of trip t that the projectile is same to twice the time to reach maximum elevation



Substitute in the above equation



The upright acceleration because that the projectile after ~ launch is same to acceleration as result of gravity g because only pressure acting on that is gravitational force.

Do not take upright component of initial velocity together

since it is same to

Determine the angle for which time of trip is more by using derived expression for time that flight.

Step 2 the 2

The expression for time of trip is given as follows:

From the over expression, the time of flight for projectile is directly proportional come sine value of edge of projection. Thus, for big angle time invested is large.

Therefore, projectile released with

spent an ext time in air.

The projectile fired through angle spent more time in air.

The worth of sine angle rises with boost in angle. Thus, the sine value better for edge 보다 that of angle Therefore, time of flight is an ext for angle 보다 for angle

Do no take the expression because that time of flight as since the exactly expression is .


The projectile fired v angle spent an ext time in air.

Answer only

The projectile fired v angle spent much more time in air.

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v=v, +at
v = v, sine
v=v, cos
V=0 V, sin e V, cosa
v=Voy+a, 4
v, sino
0 =v, sin 0+(-8) 4 Vo sino
4 sino
1 = 21,
sin 1=2 olg 2v, sine
v, sin 0.
2v, sino
1= sino
2v, sino
We to be unable come transcribe this image

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