A football is thrown from the edge of a cliff from a height of 22 m at a velocity of 18 m/s [39degrees above the horizontal]. A player at the bottom of the cliff is 12 m away from the base of the cliff and runs at a maximum speed of 6.0 m/s to catch the ball.
Is it possible for the player to catch the ball? Support your answer with calculations.
No. Henry's solution makes sense up until the last bit.
The player is 12m away from the base of the cliff. He's running away from the cliff, or in other words, in the same horizontal direction as the football. Therefore, you need to sub in the time (3.56s) into the equation for the horizontal velocity, to determine the range of the football:
Vh(orXo)=d/t
d=Vh*t
d=13.99*3.5672
d=49.900
Okay, so the player needs to cover a distance of 50m, in order to catch the ball.
V=6.0m/s
t=3.56s
d=?
How far can he travel in 3.56s?
d=vt
d=6*3.56
d=21.4m
21.4m+12m (he was already displaced by this much)
Therefore, the football will travel about 50m, while the player can only run a maximum of 33m. He won't be able to make it.
Well, let's clown around with some calculations to find out if our player can catch that adventurous ball!
First, we'll split the motion of the football into horizontal and vertical components. The horizontal component doesn't really matter because our player is already positioned right underneath the cliff. So let's focus on the vertical motion.
We have the initial height of the ball (h = 22 m) and the initial vertical velocity (v₀ = 18 m/s * sin(39°)). Using the kinematic equation, h = v₀t + (1/2)gt², we can solve for the time it takes for the ball to hit the ground.
Putting it into the equation, we get: 0 = (18 m/s * sin(39°))t + (1/2)(9.8 m/s²)t².
Simplifying, we have: 0 = 9.8t² + (10.42)t - 22.
Applying the quadratic formula, we find two values for t: t ≈ 0.911 s and t ≈ -2.42 s. Since time can't be negative, we'll discard the second solution.
Now, let's see how far our player can run in 0.911 seconds. The player's speed is 6.0 m/s, and multiplied by the time gives us a distance of 6.0 m/s * 0.911 s = 5.466 m.
Hooray for our player! They only need to run a distance of 5.466 m to catch the ball, and they are positioned 12 m away. So yes, it's possible for the player to catch the ball! We hope they have good sprinting shoes and a sense of humor to make it more fun! Good luck!
To determine if the player can catch the ball, we need to compare the time it takes for the ball to land at the bottom of the cliff with the time it takes for the player to reach the same point.
1. Calculate the time it takes for the ball to reach the bottom of the cliff:
First, we need to find the vertical component of the initial velocity. We can use the equation:
Vy = V * sin(θ)
where Vy is the vertical component of the velocity, V is the initial velocity (18 m/s), and θ is the angle above the horizontal (39 degrees).
Vy = 18 * sin(39)
Vy ≈ 11 m/s
Next, we can use the equation for displacement in vertical motion:
Δy = Vyo * t - 0.5 * g * t^2
where Δy is the vertical displacement (22 m), Vyo is the initial vertical velocity (11 m/s), t is the time, and g is the acceleration due to gravity (9.8 m/s^2).
Rearranging the equation and solving for t, we get:
0 = 11t - 0.5 * 9.8 * t^2
This is a quadratic equation in the form at^2 + bt + c = 0, where
a = -0.5 * 9.8 = -4.9
b = 11
c = 0
Using the quadratic formula, t = (-b ± √(b^2 - 4ac))/(2a), we can solve for t:
t = (-11 ± √(11^2 - 4*(-4.9)*0))/(2*(-4.9))
t ≈ 2.22 s
Therefore, it takes approximately 2.22 seconds for the ball to reach the bottom of the cliff.
2. Calculate the time it takes for the player to reach the same point:
The player runs at a maximum speed of 6.0 m/s to catch the ball, and the distance from the player to the base of the cliff is 12 m. Using the formula:
time = distance / speed
we can calculate the time it takes for the player to reach the bottom of the cliff:
time = 12 m / 6.0 m/s
time = 2.0 s
Therefore, it takes 2.0 seconds for the player to reach the bottom of the cliff.
Comparing the times, we see that the ball takes approximately 2.22 seconds to reach the bottom of the cliff, while the player takes 2.0 seconds to reach the same point. Since the player reaches the bottom of the cliff before the ball, it is indeed possible for the player to catch the ball.
To determine if the player can catch the ball, we need to compare the horizontal distance the ball travels with the distance the player can cover in the same time.
First, let's find the time it takes for the ball to reach the ground. We can use the vertical motion equation:
Δy = V₀y * t + 0.5 * a * t²
Since the ball is thrown horizontally, its initial vertical velocity (V₀y) is 0 m/s, and the acceleration due to gravity (a) is approximately 9.8 m/s². The change in height (Δy) is -22 m (negative because the ball is falling).
-22 = 0 * t + 0.5 * 9.8 * t²
-22 = 4.9 * t²
t² = -22 / -4.9
t² = 4.489795918
t ≈ √4.489795918
t ≈ 2.12 seconds
Now, let's find the horizontal distance the ball travels using its initial horizontal velocity (V₀x) and the time it takes to reach the ground:
V₀x = V₀ * cos(θ) (where θ is the angle above the horizontal, given as 39 degrees)
V₀x = 18 m/s * cos(39°)
V₀x ≈ 18 m/s * 0.766 = 13.788 m/s
Distance = V₀x * t = 13.788 m/s * 2.12 s
Distance ≈ 29.25 meters
The ball travels approximately 29.25 meters horizontally before hitting the ground.
Now, let's see if the player can cover that distance in the same time. The player's maximum speed is 6 m/s, and the time taken for the ball to reach the ground is 2.12 seconds.
Distance covered by the player = Speed * Time = 6 m/s * 2.12 s
Distance ≈ 12.72 meters
Since the player can only cover a distance of approximately 12.72 meters, which is less than the distance the ball travels (29.25 meters) before hitting the ground, it is not possible for the player to catch the ball.
Vo = 18m/s[39o]
Xo = 18*cos39 = 13.99 m/s.
Yo = 18*sin39 = 11.33 m/s.
Y = Yo + g*Tr = 0 @ max ht.
11.33 - 9.8*Tr = 0
9.8Tr = 11.33
Tr = 1.16 s. = Rise time.
hmax = ho + (Y^2-Yo^2)/2g
hmax = 22 + (0-11.33^2)/-19.6 = 28.55 m.
Above gnd.
h = 0.5g*t^2 = 28.55 m.
4.9*t^2 = 28.55
t^2 = 5.83
Tf = 2.41 s = Fall time.
V * (1.16+2.41) = 12
3.57V = 12
V = 3.36 m/s = Required velocity.
Yes.