A shotputter throws the shot with an initial speed of 15.5 m/s at a 32.0 degree angle to the horizontal. Calculate the horizontal distance traveled by the shot if it leaves the athlete's hand at a height of 2.20 m above the ground.
24.1 m
To calculate the horizontal distance traveled by the shot, we can break down the initial velocity into its horizontal and vertical components. The horizontal component of the initial velocity remains constant throughout the entire motion, while the vertical component changes due to the effect of gravity.
First, let's find the initial horizontal component of the velocity (Vx) and the initial vertical component of the velocity (Vy).
Vx = V * cos(θ)
Vy = V * sin(θ)
where V is the initial speed of the shot (15.5 m/s) and θ is the angle of projection (32.0 degrees).
Vx = 15.5 m/s * cos(32.0°) ≈ 13.08 m/s
Vy = 15.5 m/s * sin(32.0°) ≈ 8.29 m/s
Now, we can use the vertical motion equations to find the total time taken by the shot to reach the ground. We assume that the final vertical displacement (Δy) is equal to the negative initial displacement (-2.20 m), since the shot is thrown upward.
Δy = Vy * t + (1/2) * g * t^2
-2.20 m = 8.29 m/s * t + (1/2) * (-9.8 m/s^2) * t^2
By rearranging the equation and solving for t, we get a quadratic equation:
-0.5 * 9.8t^2 + 8.29t - 2.20 = 0
Using the quadratic formula, we can solve for the time taken. Let's use the positive value of t since we are interested in the time it takes for the shot to reach the ground.
t ≈ 0.37 s
Now we can use the horizontal motion equation to calculate the horizontal distance traveled by the shot:
Horizontal distance (D) = Vx * t
D = 13.08 m/s * 0.37 s ≈ 4.84 m
Therefore, the shot will travel approximately 4.84 meters horizontally before hitting the ground.
Split the initial velocity (u) into two components, vertical (uy) and horizontal (ux) using:
ux=u cos(θ) and
uy=u sin(θ)
Solving the time (t) the shot stays in the air from:
Sy=uy*t - (1/2)gt²
where Sy = vertical distance travelled = -2.2m
(reject the negative root).
The horizontal distance travelled is then
Sx=ux*t
What do you mean by where Sy = vertical distance traveled = -2.2m? Is the -2.2m from the equation above? Do I add that in to the equation?
-2.2m is the distance between the hand and the ground.
So we equate the vertical distance (from the hand) with the distance as a function of time:
-2.2=uy*t - (1/2)gt²
Solve for t.