A projectile remains in the air for 6.00 s after it is fired. The horizontal component of its velocity is 100 m/s. (a) How far did it move forward before it hit the ground? (b) How long after firing did it reach the highest point of its path? (c) What was the altitude of the highest point in its path?

Question

john_es · Answer

The total flying time of the projectile is t=6 s, and we assume a parabolic motion. In this case, the equations of motion are,

$$x=x_0+v_{0x}t \
y=y_0+v_{0y}t-g t^2/2$$
Where 
$$x_0=0\ m;\  y_0=0\ m;\ v_{0x}=100\ m/s;\ g=9.8\ m/s^2$$
For the questions,
(a) $$x=v_{0x}t=100\cdot6=600\ m$$
(b) Assuming a parabolic motion, and the initial conditions early metnioned, the highest point is reached at t=6/2=3 s.
(c) First, we need the value of the initial velocity in the vertical direction. So, in the complete movement, when the projectile goes from y=0 yo y=0, we have,
$$y=y_0+v_{0y}t-g t^2/2\Rightarrow 0=0+v_{0y}t-gt^2/2\Rightarrow v_{0y}=gt/2=9.8\cdot6/2=29.4\ m/s$$
The highest point is reached at t=3 s, so its value is,
$$y=v_{0y}t-gt^2/2\Rightarrow 29.4\cdot3-9.8\cdot3^2/2=44.1\ m$$

anonymous · Answer

If i had to say it would have traveled 600 meters total over the 6 seconds giving the trajectory it's highest peak would be 300 meter giving if the trajectory was an arc the speed at which it reach it highest point would be 3 seconds