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2 years ago

Which statement about projectile motion is true?

Physics

2 answers:

Shalnov [3]2 years ago

4 0

Answer:

D. The hang time of a projectile is greatest for a vertical launch.

Explanation:

As we know that range of projectile is given by the formula

$R = \frac{v^2sin(2\theta)}{g}$

so here as we increase the angle till 45 degree then the range will decrease but if we increase the angle more than 45 degree then the range will start decreasing

Also we know that time of flight is given by the formula

$T = \frac{2vsin\theta}{g}$

since it depends on the vertical component of the motion so we can say that hang time can be determine by vertical components of the motion

So maximum hang time is possible only when angle of projection is 90 degree

so correct answer will be

D. The hang time of a projectile is greatest for a vertical launch.

Svetllana [295]2 years ago

3 0

C. The range of a projectile increases with an increase in the angle of launch.

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(Please give a numerical answer, 50 points) A projectile launcher is placed on the floor and pointed straight up (90 degrees), T

Andrew [12]

Answer:

Speed at which the ball leaves the launcher = 6.34 m/s

Explanation:

The readings of height above the floor reached by the ball = 2.32, 2.26 and 2.37

Mean value $=\frac{2.32+2.26+2.37}{3} =2.317 m$

Initial height of launcher = 27 cm = 0.27 cm

So maximum height of projection = 2.317-0.27 = 2.047 meter

We have equation of motion, $v^2=u^2+2as$ , where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

In this case we need to consider only vertical factors.

Vertical displacement = 2.047 m, acceleration = acceleration due to gravity = $-9.81m/s^2$ , final velocity = 0 m/s, we need to calculate initial velocity.

$0^2=u^2-2*9.81*2.047\\ \\ u=\sqrt{2*9.81*2.047} \\ \\ u=6.34 m/s$

So, speed at which the ball leaves the launcher = 6.34 m/s

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