With what initial speed does he throw his first ball?

In summary, the conversation discusses two problems: the first involving a juggler throwing two balls in a room with a ceiling at height h, and the second involving a football kicker trying to score a field goal at a distance L and height h above the ground. The juggler's first ball is thrown vertically upwards with an initial speed of sqrt(20h) m/s and takes t= -0.1v seconds to reach the ceiling. The two balls pass each other when the second ball is thrown, at a time of t=sqrt(v0/g) seconds. The balls are h/2 meters above the juggler's hands when they pass each other. For the second problem, the relationship between the elevation angle, L, and
  • #1
mfk_1868
21
0
1-Balls thrown by a juggler: A juggler performs in a room whose ceiling is at a height h
above the level of his hands. He throws his first ball vertically upwards so that it just
reaches the ceiling. At the instant when this happens, he throws his second ball upward
with the same initial speed.
(a) With what initial speed does he throw his first ball?
(b) How much time is required for this ball to reach the ceiling?
(c) How long a time after the second ball is thrown do the two balls pass each other?
(d) How far above the juggler’s hands are the balls when they pass each other?
2-A football kicker can give the ball an initial speed of v . Within what two elevation angles
must he kick the ball to score a field goal from a point at a distance L to the front of
goalposts whose horizontal bars is at a height h above the ground?
(Hint use sin^2θ + cos^2θ = 1 to get a relation between 2 tan^2θ and
1/cos^2θ substitute
and then solve the quadratic equation)
 
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  • #2
1-(a) mechanical energy theorem implies:
Em1=Em2 (1:position of the ball when it leaves the hand of the man
2: when it reaches the ceiling)

(considering the hand the referential level)

Em1= epp + ec
= 0 + 1/2mv^2

Em2= epp + ec
= +mgh + 0

Em1 = Em2

--> 1/2mv^2 = mgh
--> 1/2v^2 = gh
--> v^2 = 2gh (considering g=10 m/s^2)
--> v^2 = 20h
--> v = squared root of (20h) m/s

(b) in this case we have a varied motion for the ball,
before it reaches the cieling it is a decelerated motion,
so the equation is:
z= -1/2gt^2 +vt +v0 ( negative because it's in the opposite way of the mass force
considering an ascending axiz OZ)

z=h
when -gt + v= 0
--> t = -v/g
--> t = -0.1v (s)

(c) when the second ball leaves the hand of the man
its equation is z= -1/2gt^2 +vt +v0

when the first ball leaves the cieling
its equation is z= 1/2gt^2 +vt

they will pass each other when
z=z

-1/2gt^2 +v0 = 1/2gt^2
--> gt^2 = v0
--> t= squared root of (v0/g) (s)

(d) because it is the same initial speed
and considering that the resistance of air is zero.
we can conclude that the balls are = h/2 m (higher than the juggler)
 
  • #3
i don't know if it is correct..
i hope that a mentor can correct it to see if there are mistakes :)
 
  • #4
for b>
check the sign of t again ...

for c>
is that the total time ? ... ;) u just missed a tiny bit , but rest of the idea is good enough

for d>
i haven't checked the calculations but why be unsure?
can u use the result u got in c?

-- AI
p.S > mfk, what have u tried for question 2? do post ur working because u can be helped better that way ... the question is straightforward and shouldn't be hard basically ...
 
  • #5
Before I "correct" anything, I would like to see what mfk_1868 has tried on this problem!
 
  • #6
i solved these problems but want to see if i solved correctly
 
  • #7
1-)
V/g=t1 V=t1*g 1/2 g*t1^2=h /sqrt {2h/g}=V =>Answer of a)
t1 = v/g /frac{/sqrt {2h/g}} {g}=t1 =>Answer of b)
t2 is the time described at c.
v*t2 - 1/2 g*t2^2 + 1/2 g*t2^2 = h v*t2=h h/v=t2
t2=/frac{h}{/sqrt {2h/g}} =>Answer of c)

h - 1/2 g t2^2 =>Answer of d)

i am not very sure about second problem
 
  • #8
mfk,
whatever u have done with question 2 .. be it wrong or right ..
do post it ... that way u get ur doubts cleared better ...

Some initial hints :
1> forget the hint the question has given for now
2> Can u come up with a equation that relates the elevation angle (say theta) , L and h?

-- AI
 
  • #9
i ll post the second question soon but is my first question true?
 
  • #10
mfk,
well i could not understand much of your last post
but if u have done the same thing A_I_ has done then u are on the right track ...
i gave a few corrections to his solutions in my last post

-- AI
 
  • #11
thanks i ll compare the solutions
 

1. What is the initial speed of the ball when it is thrown?

The initial speed of the ball when it is thrown depends on various factors such as the force applied by the person throwing the ball, the weight of the ball, and any external factors like wind resistance. It can also be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

2. How is the initial speed of the ball measured?

The initial speed of the ball can be measured using a stopwatch or a speedometer. The person throwing the ball can also estimate the speed based on their own physical capabilities and the distance the ball travels.

3. Does the initial speed of the ball affect its trajectory?

Yes, the initial speed of the ball does affect its trajectory. The greater the initial speed, the higher the ball will travel and the further it will go. The angle at which the ball is thrown also plays a role in determining its trajectory.

4. How can the initial speed of the ball be increased?

The initial speed of the ball can be increased by applying a greater force when throwing the ball, using a lighter ball, or reducing wind resistance by throwing the ball in a closed environment or using a more aerodynamic design.

5. What is the average initial speed of a thrown ball?

The average initial speed of a thrown ball can vary depending on the type of ball and the person throwing it. However, for a person of average physical capabilities, the initial speed of a thrown ball can range from 20-40 miles per hour.

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