Suppose that a shot putter can put a shot at the worldclass speed v0 = 14.00 m/s and at a height of 2.160 m. What horizontal distance would the shot travel if the launch angle θ0 is (a) 45.00° and (b) 42.00°? The answers indicate that the angle of 45°, which maximizes the range of projectile motion, does not maximize the horizontal distance when the launch and landing are at different heights.
a. Vo = 14m/s[45o]
Xo = 14*cos45 = 9.9 m/s.
Yo = 14*sin45 = 9.9 m/s.
Y = Yo + g*t = 0
Tr=Yo/g = 9.9/9.8 = 1.01 s. = Rise time.
hmax = ho + Yo*t + 0.5g*t^2
hmax=2.16 + 9.9*1.01 - 4.9*1.01^2=7.16m
Above gnd.
h = 0.5g*t^2 = 7.16
4.9t^2 = 7.16
t^2 = 1.46
Tf = 1.21 s. = Fall time.
D = Xo * (Tr+Tf)=9.9 * (1.01+1.21=21.9m
= Hor. distance.
To find the horizontal distance traveled by the shot, we can use the range formula for projectile motion, which is given by:
R = (v0^2 * sin(2θ0)) / g
where:
R is the horizontal distance traveled
v0 is the initial velocity of the projectile
θ0 is the launch angle
g is the acceleration due to gravity (approximately 9.8 m/s^2)
(a) Launch angle θ0 = 45.00°:
Plugging in the values into the formula, we have:
R = (14.00 m/s)^2 * sin(2 * 45.00°) / 9.8 m/s^2
R = (14.00)^2 * sin(90.00°) / 9.8
R = (14.00)^2 * 1 / 9.8
R ≈ 20.40816327 m
Therefore, the horizontal distance traveled when the launch angle is 45.00° is approximately 20.41 m.
(b) Launch angle θ0 = 42.00°:
Using the same formula, we have:
R = (14.00 m/s)^2 * sin(2 * 42.00°) / 9.8 m/s^2
R = (14.00)^2 * sin(84.00°) / 9.8
R = (14.00)^2 * 0.9781476 / 9.8
R ≈ 19.95918367 m
Therefore, the horizontal distance traveled when the launch angle is 42.00° is approximately 19.96 m.
From these calculations, we can observe that for the given conditions, the maximum horizontal distance is achieved when the launch angle is 45.00°, as expected. However, in this particular scenario where the launch and landing heights are different, the maximum range angle of 45.00° does not maximize the horizontal distance.
To find the horizontal distance traveled by the shot putter, we can use the equations of projectile motion. Let's break it down step by step:
Step 1: Determine the vertical and horizontal components of the initial velocity.
The initial velocity of the shot putter can be broken down into its vertical (Vy0) and horizontal (Vx0) components. Since the launch angle θ0 is given, we can use trigonometry to find these components:
Vy0 = v0 * sin(θ0)
Vx0 = v0 * cos(θ0)
Step 2: Calculate the time of flight (total time in the air).
The time of flight (T) is the total time that the shot putter remains in the air. It can be calculated using the vertical component of the velocity:
T = (2 * Vy0) / g
Where 'g' is the acceleration due to gravity (approximately 9.8 m/s²).
Step 3: Calculate the horizontal distance (range).
The horizontal distance (R) is the distance traveled by the shot putter in the horizontal direction. It can be calculated using the horizontal component of the velocity and the time of flight:
R = Vx0 * T
Now we can follow these steps to find the answers to the given angles:
(a) Launch angle θ0 = 45.00°:
Using the given angle of 45.00°, we can plug it into the equations above to find the values for Vx0 and Vy0. Then, calculate T and finally plug the values into the formula for R to find the horizontal distance.
(b) Launch angle θ0 = 42.00°:
Using the given angle of 42.00°, follow the same steps as case (a) to find the horizontal distance.
By comparing the results for both angles, we can see that the launch angle of 45.00°, which maximizes the range in projectile motion, may not maximize the horizontal distance traveled when the launch and landing are at different heights.