A ball is thrown with a speed of 100m/s attain the height of 150m, (take g=10m/s^2). calculate... (i)Time of flight (ii)Angle of projection (iii)Range???
u = speed in horizontal direction until crash into ground
v = speed up
Vi = initial speed up
A = throw angle up from horizontal in degrees
Vertical problem:
v = Vi - 10 t
at top, v = 0
so 0 = Vi - 10 t
t = Vi/10 ( time to top is t )
h = 0 + Vi t - 5 t^2 (h is height, 5 is g/2)
then at top when t = Vi/10 :
h = Vi (Vi/10) - 5 Vi^2/10^2 = Vi^2/20 at top
so
150 = Vi^2 / 20
Vi^2 = 3000
Vi = 55 m/s close enough
now geometry
tan A = Vi/100 = 55/100 = .55
so A = 29 degrees
u = 55 cos 29 = 48 m/s
time at u = twice time up = 2 (55/10) = 11 seconds in the air at horizontal speed 48
48 * 11 = 530 m
check my arithmetic, working fast. Have fun.
By the way, you threw a ball up 150 meters? You belong in the major leagues. I suspect that may be why it went over 5 football fields horizontally (or else I made serious arithmetic errors.)
224 miles/hour = 100 meters/second :) according to my search in the box above.
To calculate the required values of a ball thrown with a speed of 100 m/s and reaching a height of 150 m, we can use the equations of motion and some basic trigonometry. Here's how you can find each of these values:
(i) Time of Flight:
The ball's time of flight is the total time it spends in the air. We need to find the time it takes for the ball to reach its maximum height and then return to the ground.
Using the equation of motion:
\(h = u \cdot t + \frac{1}{2} g \cdot t^2\), where \(h\) is the height, \(u\) is the initial vertical velocity, \(t\) is the time of flight, and \(g\) is the acceleration due to gravity.
Since we know the initial vertical velocity is 0 m/s at the maximum height, the equation simplifies to:
\(h = \frac{1}{2} g \cdot t^2\)
Rearranging the equation and substituting the known values:
\(t^2 = \frac{2h}{g}\)
\(t = \sqrt{\frac{2h}{g}}\)
Plugging in \(h = 150 m\) and \(g = 10 m/s^2\):
\(t = \sqrt{\frac{2 \cdot 150}{10}} = 3 \sqrt{3}\) seconds
So, the time of flight is approximately \(3 \sqrt{3}\) seconds.
(ii) Angle of Projection:
The angle of projection is the angle at which the ball is thrown with respect to the horizontal. In this case, we need to find the angle \(\theta\).
Using the equation for the horizontal range of a projectile:
\(R = \frac{u^2 \sin(2\theta)}{g}\), where \(R\) is the horizontal range.
Since we know the initial velocity \(u = 100 m/s\) and the range \(R\), we can solve for \(\theta\):
Rearranging the equation:
\(\sin(2\theta) = \frac{R \cdot g}{u^2}\)
\(\sin(2\theta) = \frac{R \cdot 10}{100^2}\)
\(\sin(2\theta) = \frac{R}{1000}\)
To find the value of \(\theta\), we need to use the inverse sine function (sin⁻¹):
\(2\theta = \sin⁻¹\left(\frac{R}{1000}\right)\)
\(\theta = \frac{1}{2} \sin⁻¹\left(\frac{R}{1000}\right)\)
Since we don't have the value of \(R\) yet, we'll calculate it in the next step.
(iii) Range:
The range of a projectile is the horizontal distance it travels. We can find the range using the equation provided above.
Substituting the known values:
\(R = \frac{(100)^2 \sin(2\theta)}{10}\)
\(R = \frac{10000 \sin(2\theta)}{10}\)
\(R = 1000 \sin(2\theta)\)
Now that we have an equation for the range in terms of theta, we can substitute it in the equation for theta we derived earlier:
\(\theta = \frac{1}{2} \sin⁻¹\left(\frac{1000 \sin(2\theta)}{1000}\right)\)
Unfortunately, there is no simple algebraic way to solve this equation. However, you can use numerical methods or approximation techniques (such as trial and error) to find the value of \(\theta\).
To summarize:
(i) The time of flight is approximately \(3 \sqrt{3}\) seconds.
(ii) The angle of projection (\(\theta\)) can be found by solving the equation \(\theta = \frac{1}{2} \sin⁻¹\left(\frac{R}{1000}\right)\) using numerical methods or approximation techniques.
(iii) The range (\(R\)) can be found by substituting the value of \(\theta\) obtained in step (ii) into the equation \(R = 1000 \sin(2\theta)\).