Vertical Motion - gravity question?

Question

A ball is thrown 140 m upward and then falls

back to earth.

The acceleration of gravity is 9.8 m/s2. Neglecting air resistance, how long will it

be in the air? Answer in units of s.

Answer 1

ANSWER: 10,69 seconds

EXPLANATION:

the time it goes upward is equal to the time it goes downward until it reaches the grown back. All you need to calculate is the time it takes to reach the highest point (when the velocity will turn to Zero).

V = Vo - gt

S = Vot - 1/2gt(2)

Being: V = Final Velocity; Vo: Initial Velocity; g= gravity; s=altitude

Using both equations, we get this:

0 = Vo - gt; and,

140 = Vot - 1/2gt(2)

From the first equation we know that Vo = gt, substituing in the second, we then have:

140 = gt(2) - 1/2 gt(2) = 1/2 gt(2)

So, t = square root (280/g) = square root (280/9,8) = 5.345 seconds.

The total time is 2xt, so 10.69 seconds

Answer 2

For coming back,H=+1/2(9.8)t^2,t^2=28.57,t=5.345 seconds.

For going up,140=-1/2(9.8)T^2 + V(0)T

When reaches the height,V=0,means -9.8T + V(0)=0,then V(0)=+9.8T,

then,140=-4.9T^2+9.8T^2=+4.9T^2,T^2=14...

so,the ball is 10.69 seconds in air.

GOOD LUCK on your studies.

Answer 3

s=(1/2)*g*t^2

t=sqr(2*140/9.8) = 5.345 s "time up"

"time up and down"

2*5.345 = 10.6904 seconds "answer"