A ball is dropping from the roof and the window of 3m where a boy observe ball cross window in 0.5sec find the distance of wall above the window
let h be the distance to top of window.
h=1/2 g t^2 or
t=sqrt(2h/g)
but
h+3=1/2 g (t+.5)^2
h+3=1/2 g (t^2+t+.25)
h+3=1/2 g (2h/g+sqrt(2h/g)+.25)
now do your algebra, put this in the form of a quadratic and solve for h. Yes, it will take a pad of paper of algebra.
To find the distance of the wall above the window, we need to determine how long it took for the ball to reach the window.
We know that the distance the ball needs to travel from the roof to the window is 3m. Additionally, we know the time it took for the ball to cross the window, which is 0.5 seconds.
To find the time it took for the ball to reach the window, we can use the equations of motion.
The equation we can use is:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity (in this case, the ball is assumed to be dropped so the initial velocity is 0 m/s)
t = time taken
a = acceleration due to gravity (approximately 9.8 m/s^2)
Plugging in the given values, we get:
3 = 0 + (1/2)(9.8)(t^2)
Next, we can solve for t:
3 = 4.9t^2
t^2 = 3/4.9
t^2 = 0.6122
t = √(0.6122)
t ≈ 0.7828 seconds
Therefore, it took approximately 0.7828 seconds for the ball to reach the window.
Now, to find the distance of the wall above the window, we can calculate the distance the ball traveled during this time.
Using the equation of motion:
s = ut + (1/2)at^2
where:
s = distance
u = initial velocity (0 m/s)
t = time taken (0.7828 seconds)
a = acceleration due to gravity (9.8 m/s^2)
Plugging in the values, we get:
s = 0 + (1/2)(9.8)(0.7828^2)
s = 0 + (1/2)(9.8)(0.6115)
s = 0 + 2.86
s = 2.86 meters
Therefore, the distance of the wall above the window is approximately 2.86 meters.