A ball is thrown straight downward with a speed of 0.50 m/s from a height of 4.0 meters. What is the speed of the ball 0.70 seconds after it is released?

Question

GIVEN:
Vi = 0.50 m/s
Vf = ?
d = 4 m
t = 0.70 s

Answer 1

Use the kinematics equation:
Vf= Vi+at
(where a is acceleration due to gravity: -9.8)
Then simply plug in....

Vf=(0.50)+(-9.8)*(0.70)
Vf= -6.36 m/s

Answer 2

Are you sure? This is a regents question and the answer is 7.4 m/s.

Answer 3

vf=vi+gt
= -.50-9.8*.7= -7.4m/s the negative sign indicates direction.

Answer 4

Your right when you state a = - 9.8 because the ball is thrown straight downward. But the height is not the distance.

d = vit + 1/2 at^2
d = -2.401

But to find the speed from then since the distance is negative would i use:
Vf^2 = Vi^2 + 2ad or some other equation.

Answer 5

Oh k, thanks a lot.

Answer 6

The distance cannot be negative. The answer should be 7.367 m/s because the acceleration is not negitive 9.81 m/s squared it is positive 9.81m/s squared

Answer 7

the is 5.85 you didn't square the .50

Answer 8

Ur all wrong

Answer 9

Draw a speed v against time graph by guessing of the values of your trip to school this morning. show how this graph can be used to determine distance

Answer 10

V0 = 0.50 m/s, Vx = find, d = ?, t = 0.70 s, a = 9.81 m/s^2
Vx = V0 + at
Vx = (0.50) + (9.81)(0.70)
Vx = 7.4 m/s (with sig figs)

Answer 11

Its 7.4 m/s