Question

(7-10) A golf ball is hit horizontally at 40 m/s off of a 100 m tall building (don't try this). The trajectory of the ball is shown. Use g = 10 m/s2 for this problem and neglect air resistance.

9. What speed is the ball traveling at after 3 seconds?

a) 20 m/s b) 30 m/s c) 40 m/s d) 50 m/s e) 60 m/s

10. What is the magnitude of the ball's acceleration after 3 seconds?

a) 0 m/s2 b) 3.3 m/s2 c) 6.7 m/s2 d) 10 m/s2 e) none of the above is correct

Answer 1

Given

9. Initial velocity u= 40 m/s (along horizontal)

Acceleration g= 10 m/s^2

Concept: horizontal velocity will remain 40 m/s as there is no acceleration in that direction but vertical velocity will increase from zero due to g downward.

So vertical velocity after 3 second is calculated by formula

v = u + at

v = 0 + 10*3 = 30 m/s

So speed after 3 s is resultant of 40m/s (horizontal) and 30 m/s (vertical) , i.e. V = squareroot of ( 30^2 +40^2)

V = sqrt (900+1600)

V = sqrt(2500)

V= 50 m/s

Option D is correct.

10. There is no acceleration in horizontal direction and acceleration along vertical is always g=10 m/s^2 (downward )

So net acceleration of ball at ANY Time is always equal to g (i.e10)

So acceleration after 3 seconds is 10 m/s2.

Option D is correct.