Projectile fired - find maximum height

[kurtlau]

(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?

 

[Integral]

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.

 

[M98Ranger]

(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 =