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, 2.00 m. If v₀ = 7.90 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

Given,

H = 2 m ; v0 = 7.9 m/s

we need to determine the horizontal distance for the given angles:

(a)when = 0 deg

v0x = v0 cos = v0 cos0 = 7.9 m/s and v0y = v0 sin = v0 sin0 = 0

ay = g = 9.8 m/s² and ax = 0

we know that, H = v0yt + 1/2 g t²   (but v0y = 0)

4.9 t² = 2

t = 0.64 sec

Now horizontal distance is

x = v0x x t = 7.9m/s x 0.64 = 5.06 m

Hence, when = 0 deg, x = 5.06 m

(b)when, = 40 deg

v0x = v0cos = v0 cos40 = 7.9 x cos40 = 6.05 m/s and v0y = v0 sin = 7.9 sin40 = 5.08 m/s

ay = g = 9.8 m/s² and ax = 0

we know that, H = v0yt + 1/2 g t²

4.9 t² +5.08t - 2 = 0

solving this quadratic equation we get, t = 0.304 sec

Now horizontal distance is

x = v0x x t = 6.05m/s x 0.304 = 1.84 m.

Hence, when = 40 deg ; x = 1.84 m.

(c) when = 45 deg.

v0x = v0 cos = 7.9 x cos45 = 5.59 m/s and v0y = v0 sin = 7.9 sin45 = 5.59 m/s

ay = g = 9.8 m/s² and ax = 0

we know that, H = v0yt + 1/2 g t²

4.9 t² +5.59t - 2 = 0

solving this quadratic equation we get, t = 0.286 sec

Now horizontal distance is

x = v0x x t = 5.59 m/s x 0.286 = 1.6 m.

Hence, when = 45, x = 1.6 m.