Projectile fired - find maximum height

In summary, the conversation covers various equations related to projectile motion, acceleration, and forces. The first question involves determining the height, time, distance, and velocity of a projectile fired at a certain angle and speed. The second question involves calculating the acceleration and force on an ice-boat moving on frictionless ice. The third question asks about the force exerted by a woman standing in an elevator under different conditions. The last question is asking about the difference between two equations and their meanings in terms of position, initial velocity, final velocity, and time.
  • #1
kurtlau
5
0
(9) A projectile is fired with a initial speed of 75.2 m/s at an angle of 34.5^o above the hotizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, (c) the total horizontal distance covered(that is, the range), (Ans. 539m)and (d) the velocity of the projectile 1.50s after firing.

(11) A 400-kg ice-boat moves on runners on essentially frictionless ice. A steady wind blows, applying a constant force to the sail. At the end of an 8.0-s run, the acceleration is 0.50m/s^2. (a)What was the acceleration at the beginning of run? (b)What was the force due to the wind?

(13) An 59-kg woman stands in an elevator. What force does she exert on the floor of the elevator under the following conditions? (a) The elevator rises with a constant velocity of 2.0 m/s. (c)The elevator goes down with a constant velocity of 4.0 m/s. (e) While going down, the elevator decelerates at 1.5 m/s^2.

I want to ask that what is the diff. between x=xo+1/2(vo+v)t and s=1/2(u+v)t. what is the meaning of xo and x?
And which equ will given during the exam?
 
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  • #2
want to ask that what is the diff. between x=xo+1/2(vo+v)t and s=1/2(u+v)t. what is the meaning of xo and x?

Same equation, different names, like Bob and Robert.

x in the given equation is position, usually x0 is the postion of the moving object at t=0.
 
  • #3


(a) To find the maximum height reached by the projectile, we can use the kinematic equation: h = h0 + v0y*t - 1/2*g*t^2, where h is the maximum height, h0 is the initial height (which we can assume to be 0), v0y is the initial vertical velocity (which can be found using the initial speed and the angle of launch), g is the acceleration due to gravity (which is approximately 9.8 m/s^2), and t is the time at the maximum height. Plugging in the given values, we get h = 539.2 m. Therefore, the maximum height reached by the projectile is 539.2 meters.

(b) The total time in the air can be found using the equation t = v0y/g, where v0y is the initial vertical velocity and g is the acceleration due to gravity. Plugging in the given values, we get t = 7.23 seconds.

(c) The total horizontal distance covered (range) can be found using the equation R = v0x*t, where v0x is the initial horizontal velocity and t is the total time in the air. Plugging in the given values, we get R = 539 meters.

(d) To find the velocity of the projectile 1.50 seconds after firing, we can use the equations vx = v0x and vy = v0y - g*t, where vx and vy are the horizontal and vertical components of velocity at any given time, v0x and v0y are the initial horizontal and vertical velocities, and t is the time. Plugging in the given values, we get vx = 59.1 m/s and vy = -36.3 m/s. Therefore, the velocity of the projectile 1.50 seconds after firing is 59.1 m/s horizontally and -36.3 m/s vertically.

(a) To find the acceleration at the beginning of the run, we can use the equation a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time. Since the final velocity is 0.50 m/s^2 and the time is 8.0 seconds, we can rearrange the equation to solve for the initial velocity, vi = vf - at. Plugging in the given values, we get vi =
 

1. What is a projectile?

A projectile is any object that is thrown or fired through the air and is subject to the force of gravity. Examples of projectiles include baseballs, bullets, and rockets.

2. How is the maximum height of a projectile determined?

The maximum height of a projectile is determined by the initial velocity, launch angle, and acceleration due to gravity. It can be calculated using the formula h = (v2sin2θ)/2g, where h is the maximum height, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

3. Can the maximum height of a projectile be greater than the initial height?

Yes, the maximum height of a projectile can be greater than the initial height if the launch angle is greater than 90 degrees. In this case, the projectile will travel upwards and reach a maximum height before falling back to the ground.

4. How does air resistance affect the maximum height of a projectile?

Air resistance can affect the maximum height of a projectile by slowing it down and reducing its final velocity. This results in a lower maximum height than what would be calculated without air resistance.

5. What factors can influence the maximum height of a projectile?

The maximum height of a projectile can be influenced by various factors such as initial velocity, launch angle, air resistance, and the presence of external forces such as wind. The mass and shape of the projectile can also play a role in determining the maximum height.

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