Question

In the shot put, a heavy lead weight—the "shot"—is given an initial velocity, starting from an initial elevation approximately equal to the shot putter's height, say, 1.96 m. If v₀ = 8.70 m/s, find the horizontal distance traveled by the shot for the following initial angles above the horizontal.

(a) θ₀ = 0°


(b) θ₀ = 40.0°


(c) θ₀ = 45.0°

Answer 1

Part A.

Given that launching angle is 0 deg, which means shot put is launched horizontally, Now

In horizontal direction, there is no acceleration, So horizontal velocity will always remain constant.

In vertical direction, ay = -g = -9.81 m/sec^2

h = vertical distance traveled by ball before reaching ground = -1.96 m

V0y = 0, since initial vertical velocity is zero in this case

Using 2nd kinematic equation:

h = V0y*t + (1/2)*ay*t^2

-1.96 = 0*t + (1/2)*(-9.81)*t^2

t = sqrt (2*1.96/9.81) = 0.632 sec

Now in this time horizontal distance traveled will be:

R = V0x*t

V0x = initial horizontal velocity = 8.70 m/sec (since shot put is thrown horizontally)

So,

R = 8.70*0.632

R = horizontal distance traveled by the shot put = 5.50 m

Part B.

Given that launching angle is 40 deg, Now

In horizontal direction, there is no acceleration, So horizontal velocity will always remain constant.

In vertical direction, ay = -g = -9.81 m/sec^2

h = vertical distance traveled by ball before reaching ground = -1.96 m

V0y = V0*sin A = 8.70*sin 40 deg = 5.59 m/sec,

Using 2nd kinematic equation:

h = V0y*t + (1/2)*ay*t^2

-1.96 = 5.59*t + (1/2)*(-9.81)*t^2

4.905*t^2 - 5.59*t - 1.96 = 0

Solving above quadratic equation and taking +ve root:

t = [5.59 + sqrt (5.59^2 + 4*4.905*1.96)]/(2*4.905) = 1.42 sec

Now in this time horizontal distance traveled will be:

R = V0x*t

V0x = initial horizontal velocity = 8.70*cos 40 deg = 6.66 m/sec

So,

R = 6.66*1.42

R = horizontal distance traveled by the shot put = 9.46 m

Part C.

Given that launching angle is 45 deg, Now

In horizontal direction, there is no acceleration, So horizontal velocity will always remain constant.

In vertical direction, ay = -g = -9.81 m/sec^2

h = vertical distance traveled by ball before reaching ground = -1.96 m

V0y = V0*sin A = 8.70*sin 45 deg = 6.15 m/sec,

Using 2nd kinematic equation:

h = V0y*t + (1/2)*ay*t^2

-1.96 = 6.15*t + (1/2)*(-9.81)*t^2

4.905*t^2 - 6.15*t - 1.96 = 0

Solving above quadratic equation and taking +ve root:

t = [6.15 + sqrt (6.15^2 + 4*4.905*1.96)]/(2*4.905) = 1.52 sec

Now in this time horizontal distance traveled will be:

R = V0x*t

V0x = initial horizontal velocity = 8.70*cos 45 deg = 6.15 m/sec

So,

R = 6.15*1.52

R = horizontal distance traveled by the shot put = 9.35 m

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