Learn formula to calculate projectile motion time of flight, height of a projectile, and range of projectile. Practice projectile motion problems with solutions and examples.
A \(v_x=4\,m/s\), \(v_y\)\( = 3 \,m/s\)
B \(v_x\)\(= 5\, m/s\), \(v_y\) \( = 8 \,m/s\)
C \(v_x\) \(= 6 \,m/s\), \(v_y\)\(= 4\, m/s\)
D \(v_x\)\(= 7 \,m/s\), \(v_y\) \( = 3\, m/s\)
Thus, \(\Delta y = 0\)
Also, \(u_y=u\;sin\,\theta\) and \(a_y=-g\)
From \(\Delta y = u_yt+\dfrac {1}{2}a_yt^2\)
\(\Rightarrow0=u\,sin\,\theta \,t-\dfrac {1}{2}gt^2\)
\(\Rightarrow t=\dfrac {2u\,sin\,\theta}{g}\)
Time of flight
\(T=\dfrac {2u\,sin\,\theta}{g}=\dfrac {2u_y}{g}\)
Initial velocity \(u_y=u\,sin\theta\)
\(a_y=-g\)
\((0)^2=u^2\,sin^2\theta-2g(H_{max})\)
\(H_{max}=\dfrac {u^2\,sin^2\theta}{2g}\)
A \(40 \,m\)
B \(30 \,m\)
C \(\dfrac {45}{4}m\)
D \(20 \,m\)
\(u_x=u\,cos(\theta-\phi)\)
\(u_y=u\,sin(\theta-\phi)\)
where \('\phi'\) is the angle at which the plane is inclined
A \(u_x=\)\(3\, m/s\), \(u_y=\) \(7\, m/s\)
B \(u_x=\) \(2 \,m/s\), \(u_y=\) \(6\, m/s\)
C \(u_x=\)\( 8 \,m/s\), \(u_y=\) \(9 \,m/s\)
D \(u_x=\) \(6 \,m/s\), \(u_y=\) \(6\sqrt 3\) \(m/s\)
Initial velocity along Y-axis, \(u_y=u\,sin\,\theta\) and
acceleration, \(a_y=-g\) when particle reaches at A.
\(\Delta y=0\)
\(\Delta y=u\,sin\,\theta\,T-\dfrac {1}{2}g(T)^2\)
T = time of flight
= \(\dfrac {2u\,sin\,\theta}{g}\)
Initial velocity along X-axis, \(u_x=u\,cos\,\theta\) and
acceleration, \(a_x=0\)
\(T=\dfrac {2u\,sin\,\theta}{g}\)
\(R=u_x\cdot T\)
\(=\dfrac {2u\,cos\,\theta \cdot u\,sin\,\theta}{g}\)
So, Range \(=\dfrac {2u\,sin\,\theta \cdot u\,cos\,\theta}{g} \)
\(R=\dfrac {2\,u_x\cdot u_y}{g}\)
or, Range \(\propto\) Product of two components of initial velocity
Initial velocity along Y-axis, \(u_y=u\,sin\,\theta\) and
acceleration, \(a_y=-g\) when particle reaches at A.
\(\Delta y=0\)
\(\Delta y=u\,sin\,\theta\,T-\dfrac {1}{2}g(T)^2\)
T = time of flight
= \(\dfrac {2u\,sin\,\theta}{g}\)
Initial velocity along X-axis, \(u_x=u\,cos\,\theta\) and
acceleration, \(a_x=0\)
\(T=\dfrac {2u\,sin\,\theta}{g}\)
\(R=u_x\cdot T\)
\(=\dfrac {2u\,cos\,\theta \cdot u\,sin\,\theta}{g}\)
\(\therefore\) Range (R) \(=\dfrac {u^2\,sin\,2\theta}{g} \)
For range to be maximum \(sin\,2\theta=1\)
\(\theta=\)45°
\(R_{max}=\dfrac {u^2}{g}\)
A 62.5 m
B 65 m
C 64 m
D 70 m
Initial velocity along Y-axis, \(u_y=0\)
\(\Delta y=-H\)
\(a_y=-g\)
as \(\Delta y=u_yt+\dfrac {1}{2}a_yt^2 \)
\(0×t+\dfrac {1}{2}×(-g)(t^2)=-H\)
\(t=\sqrt {\dfrac {2H}{g}}\)
Initial velocity along Y-axis, \(u_y=u\,sin\,\theta\) and
acceleration, \(a_y=-g\) when particle reaches at A.
\(\Delta y=0\)
\(\Delta y=u\,sin\,\theta\,T-\dfrac {1}{2}g(T)^2\)
T = time of flight
= \(\dfrac {2u\,sin\,\theta}{g}\)
Initial velocity along X-axis, \(u_x=u\,cos\,\theta\) and
acceleration, \(a_x=0\)
\(T=\dfrac {2u\,sin\,\theta}{g}\)
\(R=u_x\cdot T\)
\(=\dfrac {2u\,cos\,\theta \cdot u\,sin\,\theta}{g}\)
\(R=\dfrac {2\,u_x\cdot u_y}{g}\)
hence, we conclude that, even on changing the X and Y components of initial velocity, range remains constant.
vertical component of initial velocity, \(u_y=u\,sin\theta\)
\(\Delta Y=-H\)
\(a_y=-g\)
By third law of motion;
\(\Delta Y=u_yt+\dfrac {1}{2}a_y(t)^2\)
\(\Rightarrow\) \(-H=(u\,sin\theta)\,t-\dfrac {1}{2}gt^2\)
Solve the equation for (t) to get the time of flight.
X-component of initial velocity, \(u_x=u\,cos\,\theta\)
\(a_x=0\)
\(R=u\,cos\,\theta (t)\) ...(1)
Y-component of initial velocity, \(u_y=u\,sin\,\theta\)
\(\Delta Y=-H\)
\(a_y=-g\)
\(-H=(u\,sin\theta)t-\dfrac {1}{2}gt^2\)
A \(10\sqrt 3\) m
B \(20\sqrt 3\) m
C \(30\sqrt 3\) m
D \(10\) m