They land at the same time Save 4.C.2 Submit 4.C2 (1/1 submissions ased) Section
ID: 1773854 • Letter: T
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Explanation / Answer
4.0.2)
A cannon ball can be modelled as a projectile. Based on the motion of a projectile the given questions can be answered.
(a)No
the horizontal velocity of the projectile remains contant through out the motion of the projectile.
(b)Yes
The vertical velocity changes. it becomes zero at the highest point and then the projectile begins to accelerate down with acceleraion equal to g.
(c)The same
The time for which the projectile is airborne depends upon the vertical velocity not the horizontal velocity.
t = 2 vy/g = 2 vi sin(theta)/g
As discussed in (a)the horizontal velocity(vx = vi cos(theta) can not be changed if one need to change the horizontal velocity then the intial speed of the projectile or the angle at which it is fired has to be changed.
(d)This question is completely based on the simulation. (more information needed for this)
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