Question

A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the distance PQ? (b) How far away from the dart board is the dart released? A) (a) 8 cm;(b) 1.3 m B) (a) 22 cm;(b) 2.9 m C) (a) 18 cm;(b) 1.9 m D) (a) 12 cm;(b) 3.6 m E) (a) 11 cm;(b) 2.3 m

          A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on
a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the
distance PQ? (b) How far away from the dart board is the dart released?
A) (a) 8 cm;(b) 1.3 m
B) (a) 22 cm;(b) 2.9 m
C) (a) 18 cm;(b) 1.9 m
D) (a) 12 cm;(b) 3.6 m
E) (a) 11 cm;(b) 2.3 m

A dart is thrown horizontally with an initial speed of 10 m/s toward point P, the bull's-eye on
a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the
distance PQ? (b) How far away from the dart board is the dart released?
A) (a) 8 cm;(b) 1.3 m
B) (a) 22 cm;(b) 2.9 m
C) (a) 18 cm;(b) 1.9 m
D) (a) 12 cm;(b) 3.6 m
E) (a) 11 cm;(b) 2.3 m

Added by Jennifer M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sahil Kumar

10/11/2022

Step 1

5 * acceleration * time^2 Given: Initial velocity (u) = 0 m/s (since the dart is thrown horizontally) Time (t) = 0.19 s Acceleration (g) = 9.8 m/s^2 Distance PQ = 0 * 0.19 + 0.5 * 9.8 * 0.19^2 Distance PQ = 0 + 0.5 * 9.8 * 0.0361 Distance PQ = 0 + 0.5 * Show more…

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