Practice to find maximum height of a projectile, vertical displacement of a projectile & velocity of a particle at a given instant. Learn formulas how to calculate of projectile motion equations.
Time of flight of a particle projected along a horizontal plane
\(T=\dfrac {2\,u\,sin\theta}{g}\)
Here, t < T (Time of flight)
using \(s=ut+\dfrac {1}{2}at^2\)
\(x(t)=u_xt+\dfrac {1}{2}a_xt^2\)
\(x(t)=u_xt\)
\(x(t)=(u\,cos\,\theta)\,t\)
\(s=ut+\dfrac {1}{2}at^2\)
\(y(t)=u_yt+\dfrac {1}{2}a_yt^2\)
\(y(t)=(u\,sin\,\theta)\,t-\dfrac {1}{2}gt^2\)
The acceleration along \(x\;and \;y\) axis are shown in figure.
A \(x=10\sqrt 3\,m,\,\;y=5\,m\)
B \(x=10\,m,\,\;y=5\,\sqrt 3\,m\)
C \(x=5\,m,\,\;y=5\,m\)
D \(x=4\,m,\,\;y=3\,m\)
using \(v=u+at\)
\(v_x=u_x+a_xt\)
\(v_x\) \(=u_x+0.t\)
\(v_x=u_x\)
\(v_x=u\,cos\,\theta\)
using \(v=u+at\)
\(v_y=u_y+a_yt\)
\(v_y=u_y-g\,t\)
\(v_y=u\,sin\,\theta-g\,t\)
\(\vec v=v_x\hat i+v_y\hat j\)
\(\vec v=u_x\hat i+(u_y-g\,t)\,\hat j\)
\(|\vec v|=\sqrt {v_x^2+v_y^2}\)
A \(5\, m/s\)
B \(4\, m/s\)
C \(10\sqrt 3\, m/s\)
D \(3\, m/s\)
As we know,
Kinetic Energy = \(\dfrac {1}{2}\,m\,v^2\)
Kinetic Energy(At highest point) = \(\dfrac {1}{2}\,m\;(u\,cos\,\theta)^2\)
A \(t=\dfrac {10}{3}\;sec\)
B \(t=\dfrac {20}{3}\;sec\)
C \(t=4\;sec\)
D \(t=2\;sec\)
Time of Flight \(T=\dfrac {2\,u\,sin\,\theta}{g}\)
\(H_{max}=\dfrac {u^2\;sin^2\;\theta}{2g}\)
\(u_y=u\;sin\;\theta\)
\(a_y=-g\)
using \(s=ut+\dfrac {1}{2}\,at^2\) \((\because\,\, h=s)\)
\(h=u\;sin\;\theta \cdot t-\dfrac {1}{2}\,gt^2\)
\(gt^2-2\;u\;sin\;\theta\cdot t+h=0\)
\(t=\dfrac {2\;u\;sin\;\theta\pm \sqrt {4\;u^2\;sin^2\;\theta-4gh}}{2g}\)
Here, two values of time indicating projectile is achieving same value of height h, for both the instants.
A \(t=4\;sec,\;2\;sec\)
B \(t=5\;sec,\;8\;sec\)
C \(t=3\;sec,\;8\;sec\)
D \(t=8\;sec,\;4\;sec\)