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Question

A football is kicked at ground level with a speed of 20.0 m/s at an angle of 31.0° to the horizontal. How much later does it hit the ground?

Answer 1

s(x) = ucos(theta)t
s(y) = usin(theta)t - (1/2)g t^2
v(x) = ucos(theta)
v(y) = usin(theta) - gt, g is the acceleration due to gravity and is generally taken as 9.81ms^-2
The ball hits the ground when s(y) above is equal to 0, i.e.
20sin(31)t - 0.5*9.81*t^2 = 0,
=> 10.3t - 4.905t^2 = 0,
=> t(10.3 - 4.905t) = 0,
thus t = 0 or t = 10.3/4.905 = 2.1secs.
The 0 accounts for the time of initial projection of the football, and the second time of 2.1 secs accounts for when the football next hits the ground after it has been projected, and is hence the answer!!!!

Answer 2

The footballl has vertical and horizontal components of speed due to the kick.

horizontal = 20 cos 31 m/s

vertical = 20 sin 31 m/s

No force acts on the ball on the horizontal axis but gravity acts on it on the vertical. The position of the ball is given as a function of its constant velocity due to the kick and it's acceleration in the opposite direction due to the force of gravity.

s(t) = 20 sin (31) t - 1/2 9.8 t^2 = t(20 sin(31) - 1/2 9.8 t)

s(t) is zero (i.e. the footbal is on the ground) at time 0 and again when

t = 20 sin(31)/(1/2 9.8) = 2.102 seconds

Answer 3

You kick the ball in an angle. So you can see its velocity (the vector) having an horizontal and a vertical component. In this way, you can see the problema as a football having, at the same time, an horizontal movement (given by the horizontal velocity that`s 20*cos(30º) m/s), and a vertical movement (given by the initial vertical velocity, thats 20*sin(30º) m/s).
The horizontal movement has nothing to do with the ball hitting the ground (because is horizontal, it doesn`t make the ball fall). So the vertical movement is the one that makes the ball hit the ground. That movement is given by the initial vertical velocity. So you can see the problem as throwing up a ball with initial velocity 20*sin(30º) m/s and getting the time it takes to hit the ground. Because of the simmetry given because the ideal conditions supposed, that time is equal to twice the time it takes to reach the highest position. In other words, it takes the same time to go up, and to down. So the way to go is calculate twice the time it takes the ball to go up to the top of its movement. That can be calculated as the time it takes an object to get velocity 0, starting with an initial velocity 20sin(30º), in a straight line movement (because at top, velocity is 0).

Answer 4

Do a bit of trigonmetry to find the rate at which it is traveling upward (i.e., the speed at which its altitude is increasing).

From there, use the formula that you were given that includes the fact that gravitational acceleration is about 9.8 meters per second per second, and you'll quickly get to the answer.

Answer 5

Initial upward velocity is U sin Ó¨

When it hits the ground, its velocity is - U sin Ó¨.

Change in velocity is - U sin Ó¨ - U sin Ó¨ = - 2 U sin Ó¨

Acceleration is ( - g ) = change in velocity / time.

Time = g / change in velocity.

Time t = 2 U sin Ó¨ / 9.8

t = 2* 20 * sin31 / 9.8

t = 2.1 s

Answer 6

The intial velocity has two components.one component is parallel to ground while other is perpendicular to it.The parallel component of velocity is given by ucos31 & perpendicular component is usin31.Only perpendicular component is affected by gravity not parallel component.
Now we use the perpendicular component of velocity to find out time.when a ball is thrown up,its final velocity is 0.
final velocity,v=usin31+a*t.Here a is acceleration due to gravity=-9.8m/s^2 (negative sign is used as gravity is opposit to velocity)
therefore, 0=20*0.515-9.8*t
=> 10.3=9.8*t
from this we get time,t=1.05sec.