A projectile is launched at 45 degrees the landing position is the same as the launched position . At what other angle can the projectile be launched to achieve the same horizontal position?
Answers
Answer:
) The general equations of motion of the projectile on the x and y axis are:
where v0 is the initial velocity, is the angle with respect to the ground, and is the gravitational acceleration. We can see that the motion of the projectile is an uniform motion on the x-axis and an uniformly accelerated motion on the y-axis.
First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of . To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
that has two solutions: t=0 (beginning of the motion) and
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
with .
If we call the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
Now let's remind that so that we can rewrite the equation as
and using :
and we can see that there are two values of that satisfy the equation: and , which is the solution of our problem.
2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.
Explanation:
75°