A Shot-putter throws the shot (mass=7.3 kg) with an initial speed of 14.4 m/s at a 34 degree angle to the horizontal. calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.10 m above the ground. (neglect air resistance)
please show each step.
please show each step.
-
We need to break this problem up into several parts. First, we find the
time it the shot to return to the initial height after reaching its peak. Then,
we find the time it takes from that moment for the shot to hit the ground.
Then, we take the total time (which is the sum of those two times) and multiply
it by the constant horizontal velocity of the shot.
We know that since the shot is a projectile, it will have the same speed as
the initial velocity when it comes back down to its initial height after reaching its
peak. The velocity will be in the opposite direction with respect to the y axis,
though.
We can use the equation to find how long it takes for the shot to return to its initial height:
v = v0 + at
where:
v = current velocity
v0 = initial velocity
at = acceleration * time, where a = g = -9.8 m/s^2
We have to do this with respect to the y axis, though, so the velocity must
be multiplied by sin(theta). This is because right now we are only finding
the time it takes for the shot to go reach its peak and come down to its initial
height, which only has to do with vertical motion:
(-12.4 m/s)*sin(34º) = (12.4 m/s)*sin(34º) + (-9.8 m/s^2)*t
-16.10475562 m/s = (-9.8 m/s^2)*t
(-16.10475562 m/s) / (-9.8 m/s^2) = t
t = 1.64334241 s
Now, we find how long it takes from this point for the shot to hit the ground
The velocity used is already the y-component because we found it in the
previous segment. Therefore, sin(theta) is not applied again:
y = y0 + v0*t + 0.5*a*t^2
where y is current y position, y0 is initial y position, a = g = -9.8 m/s^2
0 m = 2.1 m + (-16.10475562)*t + 0.5*(-9.8 m/s^2)*t^2
-(4.9 m/s^2)*t^2 - 16.10475562*t + 2.1 m = 0
Use the quadratic formula to solve for t:
t = -3.412281557 s or t = 0.1255967368 s
t = 0.1255967368 s is the only positive answer, so the other one can be ignored.
The total time that the shot is in the air is:
t_total = 1.64334241 s + 0.1255967368 s
t_total = 1.768939147 s
Since the shot is a projectile, we know that the x component of the
velocity remains constant throughout its time in the air. Therefore,
we can just multiply this x-component of velocity by total time to get the
distance traveled along the x axis:
(14.4 m/s)*cos(34º)*(1.768939147 s) = 21.11784503 m
Horizontal distance traveled = 21.11784503 m
time it the shot to return to the initial height after reaching its peak. Then,
we find the time it takes from that moment for the shot to hit the ground.
Then, we take the total time (which is the sum of those two times) and multiply
it by the constant horizontal velocity of the shot.
We know that since the shot is a projectile, it will have the same speed as
the initial velocity when it comes back down to its initial height after reaching its
peak. The velocity will be in the opposite direction with respect to the y axis,
though.
We can use the equation to find how long it takes for the shot to return to its initial height:
v = v0 + at
where:
v = current velocity
v0 = initial velocity
at = acceleration * time, where a = g = -9.8 m/s^2
We have to do this with respect to the y axis, though, so the velocity must
be multiplied by sin(theta). This is because right now we are only finding
the time it takes for the shot to go reach its peak and come down to its initial
height, which only has to do with vertical motion:
(-12.4 m/s)*sin(34º) = (12.4 m/s)*sin(34º) + (-9.8 m/s^2)*t
-16.10475562 m/s = (-9.8 m/s^2)*t
(-16.10475562 m/s) / (-9.8 m/s^2) = t
t = 1.64334241 s
Now, we find how long it takes from this point for the shot to hit the ground
The velocity used is already the y-component because we found it in the
previous segment. Therefore, sin(theta) is not applied again:
y = y0 + v0*t + 0.5*a*t^2
where y is current y position, y0 is initial y position, a = g = -9.8 m/s^2
0 m = 2.1 m + (-16.10475562)*t + 0.5*(-9.8 m/s^2)*t^2
-(4.9 m/s^2)*t^2 - 16.10475562*t + 2.1 m = 0
Use the quadratic formula to solve for t:
t = -3.412281557 s or t = 0.1255967368 s
t = 0.1255967368 s is the only positive answer, so the other one can be ignored.
The total time that the shot is in the air is:
t_total = 1.64334241 s + 0.1255967368 s
t_total = 1.768939147 s
Since the shot is a projectile, we know that the x component of the
velocity remains constant throughout its time in the air. Therefore,
we can just multiply this x-component of velocity by total time to get the
distance traveled along the x axis:
(14.4 m/s)*cos(34º)*(1.768939147 s) = 21.11784503 m
Horizontal distance traveled = 21.11784503 m