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A baseball is thrown from the roof of a building of height 21.6 m with an initial velocity of magnitude 11.7 m/s and directed at an angle of 57.5 degrees above the horizontal.What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance.
Take free fall acceleration to be = 9.80 m/s^2 .

2007-10-23 10:57:08 · 3 answers · asked by Natiphy2007 1 in Science & Mathematics Physics

3 answers

As the baseball is thrown from the roof of a building of height 21.6 m with an initial velocity of magnitude 11.7 m/s , the base ball possesses kinetic energy (KE) as well as potential energy (PE)

Initial KE=(1/2)mu^2 =(1/2)m*11.7*11.7=68.445 m J

Initial PT =mgh=m*9.8*21.6= 211.68 m J

Initial total energy E= KE + PE

Initial total energy E= 68.445 m + 211.68 m =280.125 m J

Initial total energy E= 280.125 m J

When the ball strikes the ground, its final PE is zero and its final KE is (1/2)mV^2, where V is its velocity when it strikes the ground.

Final total energy E! = final KE + final PE

Final total energy E! =(1/2)mV^2 + zero

Ignoring air resistance, the total energy remains conserved

Final total energy E! = initial total energy E

(1/2)mV^2= 280.125 m J

cancelling 'm'

V^2 =2*280.125 =560.25

V = sq rt 560.25 =23.67 m/s

The speed of the ball just before it strikes the ground is 23.67 m/s

2007-10-24 06:26:34 · answer #1 · answered by ukmudgal 6 · 16 0

OK...easy.

At launch: TE = PE + KE = mgh + 1/2 mv^2; where the ball mass m, is launched from a height h - 21.6 m with an initial velocity v = 11.7 mps.

Just before impact: TE = PE + KE = 1/2 mV^2; where PE = 0 because all the total energy is kinetic just above ground level. V is the impact velocity (speed).

From the conservation of energy, we can set the two TE's equal. Thus, TE(h) = mgh + 1/2 mv^2 = 1/2 mV^2 = TE(0); so that gh + 1/2 v^2 = 1/2 V^2. Therefore V^2 = 2[gh + 1/2 v^2] and V = sqrt(2gh + v^2); where g = 9.8 m/sec^2, h = 21.6 m, and v = 11.7 mps. You can do the math.

But here's the physics. Total energy is constant at any waypoint in the path of that ball. Why? Because of the conservation of energy. Thus, whatever total energy it started with, e.g., at launch point, is the same total energy it will have in the end, e.g., near impact.

Further, in general total energy TE = KE + PE, it equals kinetic energy and potential energy. As TE travels from point to point in a path, the energies swap back and forth between kinetic and potential energy, but the total TE remains fixed. And look here, we don't even care what the launch angle was.

2007-10-23 11:13:16 · answer #2 · answered by oldprof 7 · 0 0

The principle is known as "terminal velocity", the maximum rate a falling object CAN fall in a gravity environment. The rate of fall isn't determined by wind resistance, it's determined by force of gravity (how large the mass is the object is falling toward) times how long it has been falling. Standard for Earth is 32 feet per second, per second of duration. At a point (air resistance notwithstanding) you fall at a constant speed, gaining no more actual speed (this is what skydivers count on- air resistance doesn't come into play until the chute opens) no matter how long you continue to fall after reaching max descent speed, a speed reached regardless of air resistance (resistance slows you [or a raindrop] at first, but is overcome after seconds of falling). It's a matter of mass, gravity and duration of fall. No, a raindrop probably would not injure you, but a penny dropped from the top of the Empire State Building might.

2016-03-13 05:30:57 · answer #3 · answered by Anonymous · 0 0

Lets split the question into two cases. First till the ball reaches the max height where its velocity becomes 0
and then from there.
So, now
Initial velocity u = 11.7 m/s
Final Velocity v = 0 m/s
By applying V=U+AT we get
0 = 11.7 +(-9.8) * T
=> T = 11.7/9.8

Now to find how much height the ball must have reached:
Use
S= UT + 0.5 AT^2
=> S= 11.7*(11.7/9.8) + 0.5*(- 9.8)*(11.7/9.8)^2
=> S= 28.584m

Now we have to find how much time it will take to fall freely from the highest point.
We will use s= ut + 0.5 at^2
=> (21.6+6.98) = 0 + 0.5*9.8*T^2
=>28.584 = 4.9 * T^2
=> t = 2.415s

To find the final velocity just before it strikes the ground:
As we know that the velocity of the ball at the heighest point just before it is about to change the direction is 0
U=0
a= 9.8m/s2
t= 2.415s
Applying V=U+AT
V=0+9.8*2.415
=> V= 23.67 m/s
Therefore, Final velcity is 23.67m/s

2007-10-23 11:14:18 · answer #4 · answered by That's me ... 3 · 0 0

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