2D Motion Question?

Question

A football is kicked at ground level with a speed of 19.4 m/s at an angle of 39.8° to the horizontal. How much later does it hit the ground?

How do you solve this problem?

Answer 1

a=-9.8m/s/s
v0=19.4Sin(39.8)
d=0
s0=0

d=s0+v0*t+.5a*t^2

Plug in values and solve for t
one of the answers will be t=0 which is when you first kicked it
the other answer will be the time it next is at the ground which is the flight time

Answer 2

Use vectors and trignometry.

Remember when the acceleration is constant you can decompose your motion into the y axis and x axis.

The only thing that determines when it will hit the ground is gravity which will pull only in the negative y direction.

Choose the y axis to be in the up direction and the x axis to be the rightward direction.

The initial speed, vo_y, in the y direction is related by the initial speed vo by:

vo_y = vo * sin (angle)

vo_y = (39.8 m/s) * sin (39.8)

this is your initial velocity at yo = 0 since gravity will pull it back down, you must figure out how long it will take for gravity to pull it back down.

Use this equation to figure out how long it will take:

at^2/2 + vo_y*t + yo = y

Where a is the gravitational acceleration -9.89 m/s^2, t is the total time, yo is the initial height, and y is the final height.

Subtract yo from both sides

at^2/2 + vo_y*t = y - yo

y - yo = 0, since the total distance is zero since it starts at the ground and falls back to the ground.

at^2/2 + vo_y*t = 0

Solve for time, t:

t = - 2*vo_y/a

t = - 2 (19.4 m/s)* sin(39.8)/-9.89 m/s^2

t = 2.51 seconds

Answer 3

First you have to divide the initial velocity into components:
x: 19.4cos(39.8)
y: 19.4sin(39.8)

The y component will determine when the ball will hit the ground, while the x component will determine how far the ball will move before the amount of time determined by y passes.

So, first thing to concentrate on is y:

take the gravitational acceleration g (usually taken as 9.81m/s^2), calculate the time of the ball's fall starting from an initial velocity directly up, of the y component, 19.4sin(39.8):

19.4sin(39.8)=t*9.81/2

although your question is answered by calculating the time, let us calculate the distance also, which is much simpler, once you find the time:

t*19.4cos(39.8)=d

where d is the distance the ball hits the ground. there you go!

Answer 4

At 39.8 angle the vertical component of velocity =19.4 x sin 39.8 =Vo
Vo/g =time
Time ball hits the ground =1.27 seconds(neglecting air resistance)

Answer 5

The ball has a velocity in the vertical and horizontal direction. We need to worry about the vertical velocity becuase it's vertical position will determine when it hits the ground.

Vertical velocity = 19.4sin(19.8) = 12.418

Now we can use the equation:

displacement = (Vinitial)(time) + (1/2)(acceleration)(time)^2

We are going to use this equation for the vertical aspect of the problem.

-Vertical displacement = 0, it is going back down to the ground
-Vinitial = 12.418 from above
-acceleration = -9.8 m/s/s must be negative, we are making up positive, because we made the velocity positive, which is up.
-time = t

0 = (12.418)(t) + (1/2)(-9.8)(t)^2
0 = t(12.418 + -4.9t)

t = 0
t = 2.53 seconds

there are two answers, because the vertical displacement of 0 occurs twice, right before he kicks it and when it comes back down.

Answer 6

It would probably hit the ground in a minutes. But this could be wrong because it depends on the size of the person who kicked it.