Trajectory of a Projectiles
A spherical ball was projected from a fixed point ,O, with a speed ,u, angels of elevation : 0, 10, 20, 30, 40, 50, 60, 70, 80, 90 to the horizontal ground surface. Find the value of maximum height travelled by the ball for which the horizontal range, R, greatest for the same value of u.
1)Range of projected ball.
2)Time of flight for corresponding angel of projection
APPARATUS AND MATERIALS USED:
Illustration showing how apparatus was set up:
METHOD USED TO CONDUCT EXPERIMENT:
1)The apparatus was set up as shown in the diagram by raising the trigger mechanism with one hand and using the alternate hand, to push the attached ball into the shaft, the screw on the bar was adjusted to produce an appropriate force. This remain stationary to produce a constant force for the projection. 2)The protractor built into the projectile launcher was adjusted to zero degrees. 3)The trigger was pulled , simultaneously starting the stop watch. 4)When the spherical ball hit the horizontal ground surface the stopwatch was stopped. The distance from the projector to where the ball landed was measured. 5)The time of flight was recorded together the distance (range) was recorded. 6)Steps 2-5 were repeated two more times
7)The previous process was repeated ,each for the displacement of the spherical ball at 10,20,30,40,50,60,70,80,90 degrees 8) An average for the distance an time of flight by the spherical ball was calculated. The data collected was tabulated .Graphs of Time of Flight vs Angle ,together with , Range vs Angel was plotted.
Projectile motion is the motion of an object into the air at an angel. The object in motion is referred to as a projectile ,in this experiment , the small spherical ball was a projectile. Its path is a parabola as shown below. They have two independent and simultaneous motion –a motion in the horizontal distance and motion in a vertical distance . Gravity being a vertical force affects only the vertical motion . When launched, a projectile acquires distinctly different horizontal and vertical speed. During the fight through the air , the horizontal speed of the projectile remains constant ;the vertical speed changes due to gravity.
The diagram shows the spherical ball will be projected from O , with a speed of u, at an angel α ,called the angel of projection, to the horizontal ground surface. H is the highest point reached and HM is the axis of the parabola. The ball declined to the ground at R, where OM = MR. OR is the RANGE and the time taken from O to R is the TIME OF FLIGHT. The range of the projectile is dependant on the horizontal component of its velocity, that is , it is dependant on the launch speed and angel(angel between the launch direction and the horizontal). The time of flight depends upon the vertical component of its velocity. Maximum range is obtained when the products of these two components is a maximum ,with a constant speed ,u. At the maximum height the particle will travel momentarily horizontal, that is its velocity is zero.
Since gravity is a constant the following can be used to solve projectiles: The constant speed at the projected angel ,u, = Average distance ÷ Average time of flight Since the average distance travelled by the ball is equal to the range u = Average range ÷ Average time of flight
The maximum height at the projected angel, h, = u^2 〖sin〗^(2 )α ÷ 2g where u = the constant speed at the projected angel, α = to the projected angel, g = force of gravity .
Time of flight, t, = 2u sinα ÷ g where u = constant speed at the projected angel, α = the angel of projection.
The Range ,R ,= u2 sinα
where g = gravity ,u= constant...
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