Question

The football is thrown when the player was D distance away. The ball is thrown at a 45 degree angle. What are the equation for the runner and the ball thrown. How hard does the ball need to be thrown in order to be caught. How far will the ball be thrown?

Answer 1

or you can look at picture on this website
the player ran with a constant velocity


http://aplusphysics.com/community/index.php/blog/168/entry-1016-physics-of-throwing-a-football/

Answer 2

hello there :)
for what parameter of the runner you want an eqn?

Answer 3

lets imagine that the initial velocity of the runner is u and final velocity (the velocity when he reaches to get the ball) be v okay
now the distance is given as D.. while starting off at velocity u he accelerates at an acceleration of a to reach final velocity v
now D=u*t + 1/2(a*t^2)
where t is the time

Answer 4

ay= - g ax=0
vy= -gt + v0 vy=v0
xy= -1/2gt^2+v0t+x0 xx=v0t

y component= -1/2gt^2+v0sin(45)t

x component= v0cos(45)t

this is what i got for my kinematics equations. is it right

Answer 5

that was just for the football player throwing the ball


for the runner i got

a= a
v=at+vo
x=1/2at^2+v0t+x0

Answer 6

yes you are right :)
its a projectile motio for the ball
so u need to consider the x and y
and by the way for the runner make v0=0 cos hes at rest

Answer 7

so his x component will be just x=t

Answer 8

ok it should be v=at

Answer 9

for the runner it is
v=at
s=1/2gt^2
a=a

Answer 10

yup :)

Answer 11

i have all of these equations, now how do i solve for my answer with how hard the ball needs to be thrown and at what distance will it be caught

Answer 12

i know time relates them all

Answer 13

okay.. now hard does the ball need to thrown is the first qn.. here we have to relate both the distances.. because inorder for the ball to be caught the distance covered by the ball should be equal to the distance covered by the runner

Answer 14

so the distance would have to be D=d+w
because the ball was thrown when he was at a distance d and he continued to run a distance of w

Answer 15

make the answer interms of acceleration of the ball.. how hard denotes the force.. since mass of the ball is a constant.. let us assume it as ' m'.. this accelaration multiplied by ' m' gives the force

Answer 16

the distances to be equated are .the point from where he starts running and the point from where the ball is thrown.. no other distances included

Answer 17

becuase any distance travelled before throwing the ball does not affect the force

Answer 18

how far the ball would be thrown? thats the displacement eqn for a projectile..
:)

Answer 19

i suppose because the player will only keep running until the ball stops.

Answer 20

when the ball stops he stops

Answer 21

i just figured since he ran a distance of D before the ball was thrown he also ran a distance of W to catch the ball that was thrown. so we would need D+W to calculate total distance.

Answer 22

is he running D with the ball?

Answer 23

what i meant was.. he is holding theball and running 5m forward... after 5m he throws the ball forward... lets assume the ball moves 5m forward.. so how many meters should he run?

Answer 24

No
the ball is thrown when the player is D distance away but he still travels a short distance until he catches the ball.

Answer 25

the player throwing the ball is still

Answer 26

there are 2 ppl?

Answer 27

can you explain the scenario

Answer 28

There are two people playing football
The quarter back throws the football when the runner back gets a distance of D.