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Solved University-Physics Problems

Physics 1.-  Mechanics:  Parabolic Motion 

Javier Montenegro Joo

jmj@VirtualDynamics.Org

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Parabolic Motion.-

 
 

[1] In a place where the acceleration of the gravity is 12 m/s2 , a cannon shoots projectiles with a velocity of 140 m/s. (a) Find the elevation angle  of the cannon so that the projectile hits a target at 1000 m on the same level (b) Find the flight time of the projectile (c) Compute the maximum height of the projectile.

Solution.-

For a given shooting velocity there are two complementary cannon elevation angles to hit the same target. Each of these two angles sends the projectile to different parabolic trajectories, with different flying times and different maximum heights. 

 
     
 

[2] A cannon with an elevation angle of 37o   shoots a projectile with a velocity of 115 m/s. The projectile hits the ground at a distance of 1000 m. Determine: (a) the acceleration of the gravity (b) the elevation angle of the cannon so that the projectile lands at 500m distance.

Solution.-

 
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[3] In a place where the acceleration of the gravity is 12 m/s2 ,  two cannons simultaneously shooting with a velocity of 200 m/s, are separated by a distance of 1800 m. The projectiles shot by the cannons must hit a target located at middle way between both, but at different times. Find (a) the elevation angle of each cannon (b) the flight time of each projectile (c) The time interval between impacts.

Solution.-

For a given shooting velocity, the same horizontal range can be attained with two elevation angles that are complementary, this is, that add up to 90o .

 
     
 

[4] From an elevated position at  150m  over a horizontal terrain, in a region where the Acceleration of the Gravity is  9.8m/s2 ,  a projectile is shot with an elevation angle of 40o respect to the horizontal, and a velocity of   125m/s. Determine the horizontal range.

Solution.-

The time the projectile takes to reach the ground along a parabolic trajectory -this is, the flight time- is also the time it takes to displace horizontally and, it is as well the time it takes to displace vertically. Thus the first magnitude to be calculated is the flight time.

 
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[5] From an elevated position at   181 m  over a horizontal terrain, at a region where the Acceleration of the Gravity is    7.15 m/s ,  a projectile is shot with an angle of   0    respect to the horizontal, and a velocity of   135 m/s . Determine: (a) horizontal displacement of projectile (b) its landing velocity (c) its landing angle.

Solution.- 

 Horizontal Shooting                              

Acceleration of Gravity  g =   7.15 m/s

Cannon vertical position   H = 181 m

Cannon elevation angle   A = 0

Shooting Velocity     Vo = 135 m/s

 
     
 

[6]  A projectile is shot with an elevation angle of 40, a velocity of  120 m/s,  in a place where the Acceleration of the Gravity is   9.00 m/s.

 (a) Determine: the maximum height reached by the projectile, its ascending time, its flight time and its horizontal range.

 (b) Get coordinates, velocities and the displacement vector angle with the x-axis,   after  5.00 s  of shooting.

 (c) Find how long after shooting, the projectile is at 150.00 m height and get its  position (x,y) and velocity (Vx,Vy) at that time.

  Solution.-   

 
     
     
 

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