13-17. A brass ball is shot vertically upward from the surface of an atmosphere-

Question

13-17. A brass ball is shot vertically upward from the surface of an atmosphere-free planet with an initial speed of 20.0 m/s. One second later, the ball has an instantaneous velocity in the upward direction of 15.0 m/s 13. What ls the magnitude of the acceleration due to gravity on the surface of this planet? A) 5.0 m/s 8) 9.8 m/s C) 12 m/s D) 15 m/s E) 24 m/s 14. How long does it take the ball to reach its maximum height? A) 2.0s B) 2.3s C) 4.0s D) 4.6s E) 8.0s 15. How high does the ball rise? A) 70.0 m B) 10.0 m C) 50.0 m D) 20.0 m E) 40.0 m 16. Determine the velocity of the ball when it returns to its original position. Note: assume the upward direction is positive A) +20 m/s B) -20 m/s C) +40 m/s D) -40 m/s E) zero m/s 17. How long is the ball in the air when it returns to its original position? A) 4.0s B) 4.6s C) 8.0s D) 9.2s E) 16s 18. In a race, Marcos runs 1.00 mile in 4.02 min, mounts a bicycle, and rides back to his starting point, which also the finish line, in 3.02 min. What is the magnitude of Marcos' average velocity for the race? A) zero mi/h B) 12.1 mi/h C) 14.9 mi/h D) 17.0 mi/h E) 19.9 mi/h

Explanation / Answer

As per chegg guide lines I am working first problem, kindly post remianing question in the next post

(13)

Apply kinematic equation

v= u+ gt

15= 20 + g (1)

g=- 5 m/s^2

magnitude g = 5 m/s^2

option(A) is correct answer

(14)

t= u/g

= 20/5

= 4 s

option(c) is correct answer

(15)

hmax = u^2/ 2g = ( 20)^2/ 2( 5) = 40 m

option(E) is correct answer

( 16)

v= u+ gT

= u- g( 2u/g)

= -u

= - 20 m/s

option(B) is correct answer

(17)

T= 2u/g = 2( 20)/5 = 8 s

option(c) is correct answer

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