A shot putter releases the shot some distance above the level ground with a velo

Question

A shot putter releases the shot some distance above the level ground with a velocity of 10.4 m/s, 49.2° above the horizontal. The shot hits the ground 2.06 s later. You can ignore air resistance. (Assume the horizontal direction of the shot and upward are positive.)

(a) What are the components of the shot's acceleration while in flight?

(b) What are the components of the shot's velocity at the beginning of its trajectory?

What are the components of the shot's velocity at the end of its trajectory?

(c) How far did she throw the shot horizontally?

(d) How high was the shot above the ground when she released it?

Explanation / Answer

a)

x component of the shot's acceleration = 0

y component of the shot's acceleration = g = 9.8 m/s^2

b)

x component of starting velocity = v*cos(theta) = 10.4*cos(49.2) = 6.79 m/s

y component of starting velocity = v*sin(theta) = 10.4*sin(49.2) = 7.87 m/s

there is no acceleration for horizontal motion so

x component of shot's velocity at the end of its trajectory = 6.79 m/s

let y component of shot's velocity at the end of its trajectory = V

first shot goes maximum height and then released from the maximum heigth.

maximum heigth = H

using equation

v^2 = u^2 + 2as

0 = (7.87)^2 - 2*9.8*H

H = (7.87)^2/(2*9.8) = 3.16 m

t is time taken to get max heigth so

v = u + at

0 = 7.87 - 9.8*t

t = 7.87/9.8 = 0.803 sec

time taken for rest of motion = 2.06 - 0.803 = 1.257 sec

now

V = 0 + 9.8*1.257 = 12.318 m/s

y component of shot's velocity at the end of its trajectory = 12.318 m/s

d)

now travelled distance in 1.257 sec

h = 0 + (1/2)*9.8*(1.257)^2 =

required height = h - H = 7.742 - 3.16 = 4.582 m

c)

horizontal distance = 6.79*2.06 = 13.98 m

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