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Bad White [126]
3 years ago
13

a baseball is hit 3 feet above ground level at 100 feet per second and at an angle of 45 with respect to the ground. (g=32 feet/

sec^2) find the maximum height
Physics
1 answer:
LiRa [457]3 years ago
6 0

Answer:

hmax=81ft

Explanation:

Maximum height of the object is the highest vertical position along its trajectory.

The vertical velocity is equal to 0 (Vy = 0)

0=V_{y}-g*t=v_{0}*sin(\alpha)-g*th\\

we isolate th (needed to reach the maximum height hmax)

th = \frac{v_{0}*sin(\alpha)}{g}

The formula describing vertical distance is:

y = Vy * t-g* t^{2} / 2

So, given y = hmax and t = th, we can join those two equations together:

hmax = Vy* th-g*th^{2}/2

hmax =Vo^{2}*sin(\alpha )^{2}/(2*g)

if we launch a projectile from some initial height h all you need to do is add this initial elevation

hmax =h+Vo^{2}*sin(\alpha)^{2}/(2*g)

hmax =3+100^{2}*sin(45)^{2}/(2 * 32)=81 ft

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soldi70 [24.7K]

Answer:

13 km

Explanation:

The bird flies from the runner, to the finish line, and back to the runner.  We can write two equations for the distance it travels:

d = 7.8 km + 7.8 km − 4.9 km/hr × t

d = 24.5 km/hr × t

Solve for t in the second equation and substitute into the first:

t = d / 24.5

d = 7.8 + 7.8 − 4.9 (d / 24.5)

d = 15.6 − 0.2 d

1.2 d = 15.6

d = 13

The bird flies a cumulative distance of 13 km.

4 0
3 years ago
Which has more electron shells: oxygen or sulfur? How do you know?
Sever21 [200]
Deffinitly oxygen cause everyone breathes and yah.
3 0
3 years ago
Read 2 more answers
Two identical blocks are attached to the same massless rope, which is strung around two massless, frictionless pulleys. A massle
Alik [6]

Answer:

The value of mass 1, m1= 6/5m

The value of mass 2, m2= 3/5m

Explanation:

case 1:

here tension and the acceleration will be:

for m1;

  • mg-T=ma
  • 2mg - 2T = 2ma  .....1.

for m2:

  • 2T-mg = ma/2 ..... 2.

adding the both equations,

2mg - 2T + 2T-mg = 2ma + ma/2

a = 2/5 g

putting the value of a into the equation 1.

mg - T = m* (2/5)g

T = 3/5 ( mg )

now

case 2:

The two identical blocks are released from the rest, the tension remains the same as the case 1.

so,

for m1:

  • 2T-m2g=0

for m2:

  • 2m2g - 2T =0

adding both equations we get,

2T-m2g + 2m2g - 2T = 0

m2 = m1 / 2

T = m1*g / 2

here we know that

T (case1) = T (case2)

3/5 ( mg ) = m1*g / 2

m1 = 6/5 m

hence

m2 = 3/5 m

learn more about tension here:

<u>brainly.com/question/23590078</u>

<u />

#SPJ4

4 0
2 years ago
What should a sailboat operator do when approaching a PWC head-on
nekit [7.7K]

<u><em>In accordance with the International Regulation for the prevention of collisions at sea</em></u><u>: </u>

<u>1.- A sailing boat has a passing preference over a motorized boat, </u><u>except when the motor boat is limited by its draft</u><u>. </u>

<u>2.- The sailboat must maintain its course and speed. </u>

<u>3.- </u><em><u>If it is evident that the PWC does not respond</u></em><u>, the sailboat must sound the warning signal, and change its course to starboard. </u>

<u>4.- </u><u><em>All actions must be taken as soon as possible</em></u><u>. </u>

<u>5.- If a sailboat is using its engine, the situation changes, and in that case, both ships must alter to starboard.</u>

8 0
3 years ago
A man does 1500 J of work to pull a box of 150 N. what is the displacement of the box?
Assoli18 [71]
  • Work done=1500J
  • Force=150N
  • Displacement =?

We know

\boxed{\sf Work\:Done=Force\times Displacement}

\\ \sf\longmapsto 1500=150\times Displacement

\\ \sf\longmapsto Displacement=\dfrac{1500}{150}

\\ \sf\longmapsto Displacement=10m

5 0
2 years ago
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