A soccer ball is kicked at ground level with a speed of 30 m/s at an angle of 35 degree to the horizontal. how long does it takes for the ball to hit the ground?
h = -4.9t^2 + vt + 0, where v = 30sin35°
so solve
4.9t^ - 3sin35 t = 0
t(4.9t - 3sin35) = 0
t = 0 , at the start
or
t = 30sin35/4.9 = appr 3.5 seconds
What is the vertical time? Hint: Remember it is half of horizontal time.
Step 1: Split the initial velocity into horizontal and vertical components.
The initial velocity of the ball can be split into two components: the horizontal component and the vertical component. The horizontal component remains constant throughout the motion, while the vertical component changes due to the effect of gravity.
Horizontal component (Vx): V * cos(θ)
V = 30 m/s (given)
θ = 35° (given)
Vx = 30 m/s * cos(35°) ≈ 24.46 m/s (rounded to 2 decimal places)
Step 2: Calculate the time it takes for the ball to reach the ground using the vertical component.
The time it takes for an object to hit the ground can be determined using the equation of motion in the vertical direction:
y = V0y * t - (1/2) * g * t^2
where:
y = 0 (since the ball hits the ground)
V0y = V * sin(θ)
g = acceleration due to gravity = 9.8 m/s^2 (approximately)
Using the quadratic equation, we can solve for t.
0 = (V * sin(θ)) * t - (1/2) * g * t^2
Step 3: Solve for t using the quadratic equation.
Rewriting the equation in standard quadratic form:
(1/2) * g * t^2 - (V * sin(θ)) * t = 0
Applying the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / 2a
where:
a = (1/2) * g = (1/2) * 9.8 ≈ 4.9
b = -(V * sin(θ)) = - (30 * sin(35°)) ≈ -17.06
c = 0
t = (17.06 ± sqrt((-17.06)^2 - 4 * 4.9 * 0)) / (2 * 4.9)
t = (17.06 ± sqrt(291.5236)) / 9.8
Taking the positive root:
t = (17.06 + sqrt(291.5236)) / 9.8 ≈ 3.14 seconds (rounded to 2 decimal places)
Therefore, it takes approximately 3.14 seconds for the soccer ball to hit the ground.
To find the time it takes for the soccer ball to hit the ground, we can break down its motion into horizontal and vertical components.
First, let's analyze the vertical motion of the ball. We can use the formula h = ut + (1/2)gt^2, where h is the vertical displacement (which is 0 since the ball starts and ends at ground level), u is the initial vertical velocity (which is 0 since the ball is kicked horizontally), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time.
Since we are solving for time, we can rearrange the formula to t = √(2h/g). Plugging in the values, we have t = √(2(0)/(-9.8)), which simplifies to t = 0 seconds. This means that the ball hits the ground instantly in the vertical direction.
Now, let's analyze the horizontal motion of the ball. We can use the formula s = ut, where s is the horizontal displacement, u is the initial horizontal velocity (which is 30 m/s), and t is the time.
We can rearrange the formula to solve for time, t = s/u. The horizontal displacement can be calculated using the formula s = ut + (1/2)at^2, but since there is no horizontal acceleration, this simplifies to s = ut.
Plugging in the values, we have t = s/u = 0/un = 0/(30*cos(35)), where n is the angle in radians. Simplifying further, we have t = 0/22.846 = 0 seconds.
Therefore, the ball takes 0 seconds to hit the ground horizontally as well.