A toy plane p starts flying from point A along a straight horizontal line 20 m above ground level starting with zero initial velocity and acceleration 2 m /s² as shown. At the same instant, a man P throws a ball vertically upwards with initial velocitu 'u'. Ball touches(coming to rest) the base of the plane at point B of plane's journey when it is vertically above the man. 'S' is the distance of point B from point A. Just after the contact of ball with the plane, acceleration of plane increases to 4m/s². Find:
(1) initial velocity 'u' of ball.
(2) distance 's'
(3) distance between man and plane when the man catches the ball back.(g=10 m/s²)
well, first do the vertical problem, the ball
how long does it take to fall 20 m?
20 = (1/2) (10) t^2
t^2 = 4
t = 2 seconds
so it will take to seconds to fall. By symmetry it will also take 2 seconds to rise and hit the plane
how fast is it going when it hits the ground?
u = a t
= 10*2 = 20 m/s
again by symmetry it will need 20 m/s initial velocity up to peak at 20 m
Now work on the horizontal problem, the plane.
it accelerates at 2 m/s^2 for 2 seconds
s = (1/2) a t^2 = (1/2)(2) (4)
s = 4 meters
its speed then is
v = a t = 2*2 = 4 m/s
it goes on for another 2 seconds, accelerating faster while the ball falls
d = Vi t + (1/2) a t^2
d = 4(2) + (1/2)(4)(4)
= 8 + 8
= 16
so
the horizontal distance between man and plane is 16 meters when he catches the ball
BUT
the plane is still 20 meters up
so
total distance = sqrt(16^2+20^2)
= 25.6 meters
Could have been more precise
You need to work upon it.
Genius
To solve this problem, we can break it down into three parts:
Part 1: Finding the initial velocity 'u' of the ball.
Part 2: Finding the distance 's' between points A and B.
Part 3: Finding the distance between the man and the plane when the man catches the ball back.
Let's solve each part step by step:
Part 1: Finding the initial velocity 'u' of the ball.
We know the initial velocity of the toy plane is zero, and its acceleration is 2 m/s². From this information, we can determine the time it takes for the plane to reach point B.
Using the kinematic equation: v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken,
For the plane, when it reaches point B, the final velocity is 0 m/s (since it touches the base and comes to rest). The initial velocity 'u' is 0 m/s, and the acceleration 'a' is 2 m/s².
So, 0 = 0 + 2t,
which gives us t = 0.
Therefore, it takes the toy plane zero seconds to reach point B.
Now, let's move to Part 2.
Part 2: Finding the distance 's' between points A and B.
Since the toy plane reaches point B when the ball touches it, the distance 's' is the same as the vertical distance traveled by the ball.
Using the equation for the vertical motion of the ball with constant downward acceleration due to gravity (g = 10 m/s²), we can find the time it takes for the ball to reach point B.
The equation is: s = ut + (1/2)gt²,
where s is the distance traveled, u is the initial velocity, t is the time taken, and g is the acceleration due to gravity.
We know that at point B, the final velocity of the ball is 0 m/s (because it comes to rest). So, the equation becomes: s = ut + (1/2)gt² = 0 + (1/2)(-10)(t)².
Given that t = 0 (as found in Part 1), the distance 's' is also 0.
Therefore, the distance between points A and B (s) is 0 meters.
Finally, let's move to Part 3.
Part 3: Finding the distance between the man and the plane when the man catches the ball back.
After the man throws the ball vertically upwards, it reaches its highest point and then falls back down. The man catches the ball when it is at the same height as the plane (20 meters above ground level).
To find the time it takes for the ball to reach this height, we can use the equation: v = u + gt,
where v is the final velocity (which is 0 m/s at the highest point), u is the initial velocity, g is the acceleration due to gravity, and t is the time taken.
Given that the final velocity is 0 m/s, the equation becomes: 0 = u + (-10)t.
Therefore, t = u/10.
Now, we need to find the horizontal distance traveled by the plane during this time t. The plane's initial acceleration is 2 m/s², and it increases to 4 m/s² after the ball touches it.
Using the equation: s = ut + (1/2)at²,
where s is the distance traveled, u is the initial velocity, t is the time taken, and a is the acceleration,
For the plane, the initial velocity 'u' is 0 m/s, the time 't' is u/10 (from the equation above), and the acceleration 'a' is 4 m/s².
Substituting these values into the equation, we get: s = 0 + (1/2)(4)(u/10)².
Simplifying the equation, we get: s = (2/10)(u/10)² = (1/25)u².
Finally, we need to find the distance between the man and the plane, which is the horizontal distance traveled by the plane (s) when the man catches the ball back.
Therefore, the distance between the man and the plane when the man catches the ball back is (1/25)u².
I really like your logical explanation.
Thank you so much.