Solve Shot Put Launch Angle Homework: Speed 15.1m/s

[Rhoadsb]

Homework Statement

The men's world record for the shot put, 23.12m, was set by Randy Barnes of the United States on May 20, 1990. If the shot was launched from 6.00ft above the ground at an initial angle of 42.0^o, what was its initial speed?

Homework Equations

R=2(Vo^2/g)Sin(theta)Cos(theta)
x=Voxt
Tmax= Voy/g
Ttot=2Voy/g

The Attempt at a Solution

Initially, I solved for Vo in the range equation, and I came up with 23.12=2(Vo^2/g)sin(42)cos(42)= 15.1 m/s
I know I need to break it up into x and y components, but I'm not really sure how without any Vxo, time, or Vo.

 

[anathema]

Hello,

To solve for the initial speed (Vo), we can use the horizontal range equation, which is x = Vx * t. Since we know the initial angle (42 degrees) and the initial height (6.00ft), we can use trigonometry to find the initial velocity in the x-direction (Vx). We can use the formula Vx = Vo * cos(theta), where theta is the initial angle. So, Vx = Vo * cos(42) = Vo * 0.7431.

Next, we can use the vertical range equation to solve for the time (t) it takes for the shot to reach the ground. The vertical range equation is y = Voy * t - (1/2) * g * t^2. Since the shot starts at 6.00ft above the ground and ends at ground level (0ft), we can set y = -6.00ft and solve for t. This gives us the equation -6.00 = Voy * t - (1/2) * 9.8 * t^2. We can rearrange this to t^2 - (2 * Voy / 9.8) * t + 1.2245 = 0. Using the quadratic formula, we can solve for t, which gives us two values: t = 0.215s and t = 5.677s. Since the shot is in the air for a much shorter time than 5.677s, we can assume that the correct value for t is 0.215s.

Now that we have the time (t) and the initial velocity in the x-direction (Vx), we can use the horizontal range equation to solve for the initial speed (Vo). Plugging in the values we know, we get 23.12m = Vx * t = (Vo * 0.7431) * 0.215, which gives us Vo = 44.05m/s.

Therefore, the initial speed of the shot was 44.05m/s. I hope this helps!

 

[tomcue]

I would approach this problem by first identifying the known variables and what is being asked for. The known variables are the initial angle of the shot (42.0o), the initial speed (15.1m/s), and the height from which the shot was launched (6.00ft). The unknown variable is the initial speed at which the shot was launched.

To solve for the initial speed, we can use the equation R=2(Vo^2/g)Sin(theta)Cos(theta), where R is the range, Vo is the initial speed, g is the acceleration due to gravity, and theta is the launch angle. We can rearrange this equation to solve for Vo:

Vo = √(Rg/2sin(theta)cos(theta))

Plugging in the known values, we get:

Vo = √(23.12m * 9.8m/s^2 / 2 * sin(42.0o) * cos(42.0o)) = 15.1 m/s

Therefore, the initial speed of the shot was 15.1 m/s. This is consistent with the given initial speed in the homework statement, so our solution is correct.

To break it down into x and y components, we can use the equations x=Voxt and Ttot=2Voy/g. Since we are only given the initial speed and launch angle, we cannot determine the time of flight or the x component of the velocity. However, we can calculate the y component of the velocity using the equation Tmax= Voy/g:

Tmax = 2Voy/g

Solving for Voy, we get:

Voy = g * Tmax/2 = 9.8 m/s^2 * 6.00ft / 2 = 14.7 m/s

Therefore, the initial y component of the velocity is 14.7 m/s. This can also be used to calculate the time of flight using the equation Ttot=2Voy/g.

In conclusion, the initial speed of the shot was 15.1 m/s and the initial y component of the velocity was 14.7 m/s. More information, such as the time of flight or the x component of the velocity, would be needed to fully solve this problem.