If you hadto throw one of the fourtoys above, which one would take the MOST force to throw
1 answer:
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1) The law of motion of the projectile is
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
To find the velocity, we should compute the derivative of h(t):
So now we can calculate the speed at t=2 s and t=4 s:
The negative sign in the second speed means the projectile has already reached its maximum height and it is now going downward.
2) The projectile reaches its maximum height when the speed is equal to zero:
So we have
And solving this we find
3) To find the maximum height, we take h(t) and we just replace t with the time at which the projectile reaches the maximum height, i.e. t=2.30 s:
4) The time at which the projectile hits the ground is the time at which the height is zero: h(t)=0. So, this translates into
This is a second-order equation, and if we solve it we get two solutions: the first solution is negative, so we can ignore it since it's physically meaningless; the second solution is
And this is the time at which the projectile hits the ground.
5) The velocity of the projectile when it hits the ground is the velocity at time t=4.68 s:
with negative sign, because it is directed downward.
Answer:
Explanation:
This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.
The initial velocity is in the x-direction, and there is no acceleration in the x-direction.
On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.
Applying the equations of kinematics in the x-direction gives
For the y-direction gives
Combining both equation yields the y_component of the final velocity
Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.