Practice to calculation of time of flight, maximum height, range, projection up an inclined plane, and maximum range along an Incline.
The particle has two accelerations :
Along \(x\) direction :
Along \(y\) direction :
Using,
\(s_y=u_yt+\dfrac{1}{2}\,a_yt^2\)
At time of flight \(T\)
\(s_y=0,\;t=T\)
\(0=u_y\,T+\dfrac{1}{2}a_y\,T^2\)
\(0=u\,sin\,\theta.T+\dfrac{1}{2}(-g)\,T^2\)
\(T=\dfrac{2usin\,\theta}{g}\)
\(T=\left|\dfrac{2\,u_y}{g}\right|\)
Note : The equation implies that the time of flight is independent of the acceleration along horizontal direction.
A \(4\,sec\)
B \(5\,sec\)
C \(3\,sec\)
D \(6\,sec\)
The particle has two accelerations :
Along \(x\) direction
Along \(y\) direction
Using \(v_y^2=u^2_y+2\,a_y\;s_y\)
when \(v_y=0,\;s_y=H_{max}\)
\(0=(u\,sin\,\theta)^2+2(-g)\,H_{max}\)
\(H_{max}=\dfrac{u^2sin^2\theta}{2g}\)
\(H_{max}=\left|\dfrac{u^2_y}{2\,a_y}\right|\)
A \(11.25\,m\)
B \(400\,m\)
C \(13\,m\)
D \(15\,m\)
Along \(x-\) axis
\(u_x=u\,cos\,\theta\)
\(a_x=-g\,sin\,\alpha\)
Along \(y-\) axis
\(u_y=u\,sin\,\theta\)
\(a_y=-g\,cos\,\alpha\)
using
\(s_y=u_yt+\dfrac{1}{2}\,a_yt^2\)
Let the time of flight be given by \(T\).
At \(t=T\), the displacement in \(y\) direction will be zero.
\(0=u_yT+\dfrac{1}{2}\,a_yT^2\)
\(0=u\,sin\,\theta\;T+\dfrac{1}{2}\,(-g\,cos\,\alpha)\,T^2\)
\(T=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}=\left|\dfrac{2\,u_y}{a_y}\right|\)
Let maximum height be given by \(H_{max}\).
using, \(v^2_y=u^2_y+2a_ys_y\)
\(0=u^2sin^2\theta-2g\,cos\,\alpha\;H_{max}\)
\(H_{max}=\dfrac{u^2sin^2\theta}{2g\,cos\,\alpha}=\dfrac{u^2_y}{2|a_y|}\)
A \(T=5\,sec,\;H=25\,m\)
B \(T=8\,sec,\;H=5\,m\)
C \(T=7\,sec,\;H=10\,m\)
D \(T=9\,sec,\;H=12\,m\)
Initial velocity, \(u_x=u\,cos\,\theta\)
Acceleration, \(a_x=g\,sin\,\alpha\)
Initial velocity, \(u_y=u\,sin\,\theta\)
Acceleration, \(a_y=-g\,cos\,\alpha\)
using
\(s_y=u_yt+\dfrac{1}{2}\,a_yt^2\)
At \(t=T\), the displacement in \(y\) direction will be zero.
\(0=u_yT+\dfrac{1}{2}\,a_yT^2\)
\(0=u\,sin\,\theta\;T+\dfrac{1}{2}(-g\,cos\,\alpha)\,T^2\)
\(T=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\)
\(T=\left|\dfrac{2\,u_y}{a_y}\right|\)
Let maximum height be \(H_{max}\).
using, \(v_y^2=u_y^2+2a_ys_y\)
\(0=u^2sin^2\,\theta+2(-g\,cos\,\alpha)\,H_{max}\)
\(H_{max}=\dfrac{u^2sin^2\,\theta}{2g\,cos\,\alpha}\)
\(H_{max}=\dfrac{u_y^2}{2|a_y|}\)
A \(6\,sec,\;22.5\,m\)
B \(7\,sec,\;30\,m\)
C \(8\,sec,\;25\,m\)
D \(9\,sec,\;20\,m\)
The particle has two accelerations
Along \(x\) direction
Along \(y\) direction
Using,
\(s_x=u_xt+\dfrac{1}{2}a_xt^2\)
where, \(s_x=\) Total displacement along \(x-\) axis \(=R\)
\(t=\;T=\) time of flight
\(T=\left|\dfrac{2u\,sin\,\theta}{g}\right|\)
\(R=u\,cos\,\theta\,\left(\dfrac{2u\,sin\,\theta}{g}\right)+\dfrac{1}{2}a\,\left(\dfrac{2u\,sin\,\theta}{g}\right)^2\)
\(R=\dfrac{u^2\,sin2\,\theta}{g}+\dfrac{2au^2}{g^2}\,sin^2\theta\)
A \(92\,m\)
B \(90\,m\)
C \(99\,m\)
D \(100.44\,m\)
The particle has two accelerations
Along \(x-\) direction
Along \(y-\) direction
\(R=u_x\,T+\dfrac{1}{2}a_x\,T^2\)
where, \(T=\) Time of flight
put \(T=\dfrac{2u\,sin\,\theta}{g}\)
\(R=u\,cos\,\theta\;\dfrac{2u\,sin\,\theta}{g}+\dfrac{1}{2}a_x\left(\dfrac{2u\,sin\,\theta}{g}\right)^2\)
\(R=\dfrac{u^2sin2\,\theta}{g}+2\,u^2sin^2\theta\;\dfrac{a_x}{g^2}\)
for range to be zero.
\(\Rightarrow\;\dfrac{u^2sin2\theta}{g}+2\,u^2sin^2\theta\;\dfrac{a_x}{g^2}=0\)
\(\Rightarrow\dfrac{u^2sin2\theta}{g}=-2u^2sin^2\theta\;\dfrac{a_x}{g^2}\)
\(\Rightarrow\;tan\,\theta=\dfrac{-g}{a_x}\)
This implies that acceleration \('a_x\,'\) should be along negative \(x\) axis.
A \(\dfrac{\sqrt3}{10}\ m/s^2\)
B \(\dfrac{10}{\sqrt3}\ m/s^2\)
C \(-10\sqrt3\ m/s^2\)
D \(10\ m/s^2\)
Case 1
First we take the particle projected up an inclined plane.
\(x-\) direction
Initial velocity , \(u_x=u\,cos\,\theta\)
Acceleration , \(a_x=-g\,sin\,\alpha\)
\(y-\) direction
Initial velocity , \(u_y=u\;sin\; \theta\)
Acceleration , \(a_y=-g\,cos\,\alpha\)
Here, range = OA
Motion along \(y-\) axis
When particle reaches at \(A\), displacement along \(y-\) axis becomes zero.
\(u_y=u\,sin\,\theta\)
\(a_y=-g\,cos\,\alpha\)
\(\Delta y=0\)
\(\Delta y=u_yt+\dfrac{1}{2}a_y\,t^2\)
\(0=u\,sin\,\theta\;t+\dfrac{1}{2}\,(-g\,cos\,\alpha)\,t^2\)
\(0=u\,sin\,\theta\;t-\dfrac{1}{2}\,g\,cos\,\alpha\,t^2\)
\(t=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\)
= time of flight \((T)\)
Calculation of Range \((R)\)
\(u_x=u\,cos\,\theta\)
\(a_x=-g\,sin\,\alpha\)
\(\Delta x=u_x\;t+\dfrac{1}{2}\,a_x\;t^2\)
At \(t=T\)
\(\Delta x=R=u\,cos\,\theta\;T-\dfrac{1}{2}\,g\,sin\,\alpha\,T^2\)
\(R=u\,cos\,\theta\;\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)-\dfrac{1}{2}\,g\,sin\,\alpha\,\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)^2\)
\(R=\dfrac{2u^2sin\,\theta}{g\,cos^2\,\alpha}\,cos\,(\theta+\alpha)\)
Case 2
Now the particle is projected down the inclined plane.
\(x-\) direction
Initial velocity , \(u_x=u\,cos\,\theta\)
Acceleration , \(a_x=g\,sin\,\alpha\)
\(y-\) direction
Initial velocity , \(u_y=u\,sin\,\theta\)
Acceleration , \(a_y=-g\,cos\,\alpha\)
Now range = OA
Motion along \(y-\) axis
When particle reaches at \(A\), displacement along \(y-\) axis becomes zero.
\(u_y=u\,sin\,\theta\)
\(a_y=-g\,cos\,\alpha\)
\(\Delta y=0\)
\(\Delta y=u_y\;t+\dfrac{1}{2}a_y\,t^2\)
\(0=u\,sin\,\theta. t-\dfrac{1}{2}\,(g\,cos\,\alpha)\,t^2\)
\(t=\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\)
= time of flight \((T)\)
Calculation of Range \((R)\)
\(u_x=u\,cos\,\theta\)
\(a_x=g\,sin\,\alpha\)
\(\Delta x=u_x.t+\dfrac{1}{2}\,a_x\;t^2\)
At \(t=T\)
\(\Delta x=R=u\,cos\,\theta.\;T+\dfrac{1}{2}\,g\,sin\,\alpha\,T^2\)
\(R=u\,cos\,\theta\;\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)+\dfrac{1}{2}\,g\,sin\,\alpha\,\left(\dfrac{2u\,sin\,\theta}{g\,cos\,\alpha}\right)^2\)
\(R=\dfrac{2u^2sin\,\theta}{g\,cos^2\,\alpha}\,cos\,(\alpha-\theta)\)
A \(6\,sec,\;84\,m\)
B \(5\,sec,\;225\,m\)
C \(8\,sec,\;400\,m\)
D \(9\,sec,\;100\,m\)
A \(10\,sec,\;500\,m\)
B \(12\,sec,\;400\,m\)
C \(15\,sec,\;300\,m\)
D \(17\,sec,\;600\,m\)