Question

66. Calculate the displacement and velocity at times of (a) 0.500 s, (b) 1.00 s, (c) 1.50 s, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s.

Answer 1

a) t=1s

y = 10.1m

v=5.2m/s

b) t=1.5s

y =11.475 m

v=0.3m/s

c) t=2s

y =10.4 m

v=-4.6m/s (The minus sign (-) indicates that the ball is already going down)

Step-by-step explanation:

Conceptual analysis

We apply the free fall formula for position (y) and speed (v) at any time (t).

As gravity opposes movement the sign in the equations is negative.:

y = vi*t - ½ g*t2 Equation 1

v=vit-g*t Equation 2

y: The vertical distance the ball moves at time t

vi: Initial speed

g= acceleration due to gravity

v= Speed the ball moves at time t

Known information

We know the following data:

Vi=15 m / s

`g =9.8 (m)/(s^(2) )`

t=1s ,1.5s,2s

Development of problem

We replace t in the equations (1) and (2)

a) t=1s

`y = 15*1 - ½ 9.8*1^(2)`=15-4.9=10.1m

v=15-9.8*1 =15-9.8 =5.2m/s

b) t=1.5s

`y = 15*1.5 - ½ 9.8*1.5^(2)`=22.5-11.025=11.475 m

v=15-9.8*1.5 =15-14.7=0.3m/s

c) t=2s

`y = 15*2 - ½ 9.8*2^(2)`= 30-19.6=10.4 m

v=15-9.8*2 =15-19.6=-4.6m/s (The minus sign (-) indicates that the ball is already going down)