Kinematic equation Questions Involving Velocity(final/initial),Time, Displacement, Acceleration (with gravity)

Question

Here are some questions I need help with: Please show work and explain in the best detail thanks. Please help me determine a direction for each question aswell.

1. Roget "The Rocker" Clemends can throw a baseball at 140 km/h. If Roger were to throw the ball straight up, determine the following:
a) How high will the ball go, remember that the acceleration due to gravity is 9.8 m/s^2.
b)How long will it take for the ball to reach it's highest point?
c)How long will it take for the ball to come back down?
d)How fast will it be going when it hits the ground?

2. Outside of a certain high-rise apartment building you observed a ball strike the ground at 37 m/s exactly 4.2 s after hearing someone yell "Look out Below!" Knowing that the accleration due to gravity was 9.8 m/s^2, and that each floor is 2.5 m high, which floor do you think the ball was thrown from?

I guess the acceleration due to gravity would be a direction going downward so
[D].

The following formulas are given ( d represents displacement, v represents velocity, a represents accleration, t represents time, s represents seconds, vi represents initial velocity, vf represents final velocity) : 1) d= [(vi+vf) / 2] t, 2) vf=vi+at, 3) d=vit+1/2at^2, 4) d=vft-1/2at^2, 5) vf^2=vi^2+2ad.

Answer 1

Hey, email me. There is way too much here for me to answer it all in one spot.

I can start with the first one.

Convert 140km/h to m/s...
140km/h(1000m/1km)(1h/3600s)=38.89m/s

a)V^2=Vº^2+2A(y-yº)
0=(38.89)^2+2(-9.8)(y)
0=1512.35-19.6y
19.6y=1512.35
y=77.16m above the ground.

b)aV=Vº+At
0=38.89-9.8t
9.8t=38.89
t=3.97s to reach the highest point.

c)y-yº=Vºt+(1/2)At^2

-77.16=-4.9t^2.
t^2=15.75
t=3.97s to come back down.

d)V=Vº+At
V=-9.8(3.97)
V=38.89m/s down when it hits the ground.

Answer 2

I'll answer number 1 and someone else can answer number 2.

Let's first convert 140 km/h to m/s: 140km/h(1000m/km)(1h/3600 s) = 38.89 m/s

Next do part b). At its highest point, the ball's velocity will be 0. Use the equation vf - vo = at. So 0 - 38.89 = -9.8 (t). Solving for t, you have t = 14.29 s.

Next part a). Use the equation xf - xo = vo(t) + (a(t^2)) / 2.
We want to find xf. Substituting we have xf - 0 = 38.89(14.29) + (-9.8(14.29^2)) / 2. Solving for xf, you have xf = 77.117 m.

Part c) It takes the same amount of time for the ball to return to the ground as it does for the ball to reach its apex. So multiply the time it takes to reach its highest point by 2, giving 7.936 s.

Part d) Similarly, the speed (not velocity) will be the same when the ball is thrown as when it returns to the spot at which it was thrown. So the speed will be 38.89 m/s.

Answer 3

Roget Clemends? Get a new typewriter?
Step 1. Convert 140 km/hr to m/sec. Call result vo
Step 2. Solve 0= vo -9.8t for t(top). This is answer to b.
Step 3. Solve x= vo*t(top) - 4.9 t(top)^2. This is
answer to a.
Step 4. Solve x= 4.9 t(drop)^2 for t(drop), this is answer to c.
Step 5. Solve vg= 0 + 9.8*t(drop) This is answer to d.

Prob 2. Divide 37 m/s by 9.8 to get t(drop)
Do step 4 above for the t(drop) in this problem. Get x, the distance of fall.
Divide x by 2.5 to get the estimate of floor.

Answer 4

Use the one for displacement as a function of time: x=xo + ..... Just remember that your measurement of time isn't so precise, so you're answer won't be so accurate. You can't get a better one unless you have a very tall vacuum chamber because the air resistance will become significant. If you want potential extra credit, tell your teacher that your experiment wasn't so precise, but your next best design required a very tall vacuum chamber.

Answer 5

dude this is simple physics.... read ur book. it always helps. and also draw diagrams and vector diagrams to solve.