Searches related to A football at rest is kicked so that it start to move with a velocity where horizontal component is 1/2 vm/s. In it's flight , the ball rises to a maximum height of 10m.Assuming that air resistance may be neglected and that the ground is horizontal , calculate : i) Value of v. (ii) horizontal distance travelled by the ball before striking the ground , take g =10m/s
initial vertical velocity:
at the top, vf in vertical, is zero. so in the vertical,
vf^2=vi^2+2ad
0=vi^2-2*9.8*10
vi=sqrt(2*9.8*10)
so v=sqrt(vi^2 + (1/2 vi)^2)
v=2*9.8*10 sqrt(1+1/4)
= 2*9.8*10/2 sqrt 5
time in air:
vf=vi+at
0=sqrt(20*9.8)-9.8t solve for t
then knowing t,
horizontal distance=vhor*timeinai
= 1/2 (sqrt(20*9.8))*98 sqrt5
To find the value of v and the horizontal distance traveled by the ball before striking the ground, we can use the principles of projectile motion.
Let's break down the problem step-by-step to find the answers:
Step 1: Find the initial velocity of the ball (v₀):
Since the ball is at rest before being kicked, the initial velocity (v₀) is 0 m/s.
Step 2: Find the vertical velocity (vₓ) of the ball when it reaches its maximum height:
The vertical component of the initial velocity (v₀ₓ) is 0 m/s since the ball starts at rest. At the maximum height, the vertical component of the velocity becomes 0 m/s again. Therefore, the vertical component of the velocity (vₓ) is also 0 m/s.
Step 3: Find the time taken (t) to reach the maximum height:
At the maximum height, the vertical velocity becomes 0 m/s. We can use the equation:
v = v₀ₓ + gt
Since vₓ = 0, the equation becomes:
0 = 0 + (10)(t)
0 = 10t
t = 0 seconds (since the ball is already at its maximum height)
Step 4: Find the time of flight (T):
The time of flight is the total time taken by the ball to reach the ground. It can be calculated using the equation:
T = 2t
T = 2(0)
T = 0 seconds (since the ball strikes the ground at the same instant it is kicked)
Step 5: Find the horizontal component of velocity (vₓ):
Given in the problem, the horizontal component of velocity is 1/2 vm/s. So, vₓ = 1/2 vm/s.
Step 6: Find the horizontal distance traveled by the ball before striking the ground (d):
The horizontal distance traveled by the ball can be calculated using the equation:
d = vₚ * T
Since vₚ = vₓ and T = 0, the equation becomes:
d = (1/2 vm/s) * 0
d = 0 meters (since the ball strikes the ground at the same instant it is kicked)
Therefore,
(i) The value of v is 0 m/s.
(ii) The horizontal distance traveled by the ball before striking the ground is 0 meters.
To solve this problem, we can use the principles of projectile motion. Here's how you can find the answers:
(i) Value of v:
In projectile motion, we can separate the motion into horizontal and vertical components. The initial horizontal velocity is given as 1/2 vm/s, but the initial vertical velocity is zero because the ball is initially at rest. At the maximum height, the vertical velocity becomes zero again before the ball starts to fall downward.
Using the equations of motion, we know that the vertical velocity at any given time can be calculated as:
v_vertical = u_vertical + g * t
At the maximum height, the vertical velocity is zero, so we can set this equation to zero and solve for the time taken to reach the maximum height (t_max). With g = 10 m/s²:
0 = 0 + 10 * t_max
t_max = 0 seconds
Since the horizontal velocity is constant throughout the motion and doesn't depend on the vertical component, the value of v will be the magnitude of the horizontal velocity:
v = 1/2 vm/s
(ii) Horizontal distance traveled by the ball:
To find the horizontal distance traveled by the ball, we can use the equation:
Distance = Horizontal velocity * Time
To calculate the time taken for the ball to hit the ground, we can use the equation for vertical displacement:
Displacement = Initial vertical velocity * Time + (1/2) * g * Time^2
At the highest point, the vertical displacement is 10 m, initial vertical velocity is 0 m/s, and g = 10 m/s². We can set this equation to 10 m and solve for the time of flight (t_flight):
10 = 0 * t_flight + (1/2) * 10 * t_flight^2
t_flight^2 = 2
t_flight = sqrt(2) seconds
Now we can calculate the horizontal distance traveled:
Distance = Horizontal velocity * Time
Distance = (1/2 vm/s) * sqrt(2) seconds
So, the horizontal distance traveled by the ball before striking the ground is (1/2 * sqrt(2)) vm.
Note: It's important to keep track of the units in the calculations to ensure accurate answers.