physics problem

Question

12) A ball is thrown horizontally from a 45m tall building at 7m/s.
a. What is the acceleration at 2 seconds?
b. How long does it take to reach the ground?
c. How far horizontally does it go?
d. (EXPERT) What is the final velocity the moment before it touches the ground?

Answer 1

Jessica,

Two things you have to remember about free flight at this stage of your physics class:

the only source of acceleration is gravity, and that's constant

the other is that the horizontal velocity is constant because air resistance is considered negligible.

 

a)  a = -9.8 m/s²   negative because acceleration is a vector, and the direction is down.

 

b)  y = 1/2at² + v₀t + y₀  the ground is where y = 0 and v₀ = 0 because the ball was thrown

horizontally with no vertical velocity at all. So:

 

            0 = 1/2(-9.8)t² + 45

         -45 = -4.9t²

-45/(-4.9) = t²

       3 sec = t

 

c)  If the ball was in flight for 3 seconds at 7 m/s horizontally, then:

 

       3 * 7 = 21 meters

 

d)  The overall velocity of the ball involves a vertical component and a horizontal component

     which can be combined via Pythagorean theorem. The horizontal we already know is 7.

     The vertical can be found from the equation:

 

     v = v₀ + at

 

     v = 0 + -9.8(3) = -29.4 m/s

 

     V = √((-29.4)² + (7)²) = 30.2 m/s