A guy playing basket-ball needs to make the penalty shot for his team to win. Given that the ball is thrown from a height of 2 meters at an angle of 60 degrees, that the post is situated at a distance of 7 m and standing at 3.5 m:
a) What does the initial velocity have to be?
b) What is the balls apogee?
c) What is the flight time to the basket?
I'm not sure what formulas to choose, etc. The method is not clear to me.
Since the equation of motion is
y = y0 + xtan
Since the equation of motion is
y = y0 + xtanθ - gx^2/(v cosθ)^2
we have
y = 2 + √3 x - 19.6x^2/v^2
at x=7 we have
3.5 = 2 + 7√3 - 960.4/v^2
v = 9.51 m/s
So, now our equation reads
y = 2 + √3 x - 0.217x^2
the ball reaches apogee at x = √3/.434 = 4.0 m
y = 5.456m
Since the horizontal velocity = v cosθ = 4.755m/s, it takes 7/4.755 = 1.467 seconds to fly
so it's always a tan when it's 2 dimensions? Thank you very much
typo. That's
gx^2/2(v cosθ)^2
but the numbers are right. (I hope)
To solve this problem, we can use the principles of projectile motion. Let's break it down step by step:
a) To determine the initial velocity, we can start by analyzing the vertical and horizontal components separately.
Vertical motion:
We know that the ball is thrown from a height of 2 meters and the post is standing at a height of 3.5 meters. This means that the ball needs to reach a maximum height that is 3.5 - 2 = 1.5 meters above its starting position.
Using the kinematic equation for vertical motion, we can find the time it takes for the ball to reach its highest point:
vf = vi + at
0 = vi - 9.8 * t_max
Substituting in the value of acceleration due to gravity (9.8 m/s^2) and solving for t_max, we get:
t_max = vi / 9.8
Next, we can find the initial vertical velocity (vi_y) using the relation between velocity, time, and acceleration:
vf = vi + at
0 = vi - 9.8 * t_max
0 = vi - 9.8 * (vi / 9.8)
0 = vi - vi
vi = 0
From this, we can conclude that the initial vertical velocity is zero, meaning the ball is thrown horizontally.
Horizontal motion:
The horizontal distance between the thrower and the post is 7 meters. We can use the formula for horizontal motion:
d = v * t
Since the initial vertical velocity is zero, we can conclude that the vertical time and horizontal time are the same, so t is the flight time.
Knowing that the angle between the throw and horizontal is 60 degrees, the initial horizontal velocity (vi_x) can be found using the formula:
vi_x = vi * cos(θ)
Since vi_y = 0, V = vi_x.
Now we can calculate the initial horizontal velocity (V) by substituting the given values:
V = vi_x = vi * cos(60)
b) The ball's apogee is the highest point it reaches during its flight. Since the ball is thrown horizontally, the only vertical velocity it has is due to gravity. Thus, the maximum height (h_max) can be found using the formula:
h_max = (vi_y^2) / (2 * g)
Since vi_y = 0, h_max = 0.
Therefore, the ball's apogee is 0 meters.
c) The flight time (t) can be calculated using the horizontal displacement and the horizontal velocity:
d = V * t
Rearranging the formula, we get:
t = d / V
Substituting the given values, we have:
t = 7 / V
Now you can calculate the flight time using V, which we determined in part a).
Note: It's essential to convert the angle from degrees to radians for the trigonometric functions in the formulas. To convert degrees to radians, you can use the formula:
radians = degrees * π / 180.
I hope this explanation helps you solve the problem!