The aim of this question is to utilize the equations of motion for solving 2D motion-related problems.
The speed is the rate of change of distance s with respect to time t:
v = s/t
If vf is the final speed, vi is the initial speed, a is the acceleration and s is the distance covered, the equations of motion are given by:
\[ v_{ f } \ = \ v_{ i } + a t \]
\[ S = v_{i} t + \dfrac{ 1 }{ 2 } a t^2 \]
\[ v_{ f }^2 \ = \ v_{ i }^2 + 2 a S \]
For vertical upward motion:
\[ v_{ fy } \ = \ 0, \ and \ a \ = \ -9.8 \]
For vertical downward motion:
\[ v_{ iy } \ = \ 0, \ and \ a \ = \ 9.8 \]
We will use a combination of the above constraints and equations to solve the given problem.
Expert Answer
Using the 3rd equation of motion in the vertical direction:
\[ v_{ fy }^2 \ = \ v_{ iy }^2 + 2 a S \]
Substituting values:
\[ ( 0 )^2 \ = \ v_{ iy }^2 + 2 ( -9.8 ) ( 3 ) \]
\[ \Rightarrow 0 \ = \ v_{ iy }^2 \ – \ 58.8 \]
\[ \Rightarrow v_{ iy }^2 \ = \ 58.8 \]
\[ \Rightarrow v_{ iy } \ = \ \sqrt{ 58.8 } \]
\[ \Rightarrow v_{ iy } \ = \ 7.668 m/s \]
Using second equation of motion:
\[ S = v_{iy} t + \dfrac{ 1 }{ 2 } a t^2 \]
Substituting values:
\[ 3 \ = \ ( 0 ) t + \dfrac{ 1 }{ 2 } (9.8) t^2 \]
\[ \Rightarrow 3 \ = \ 4.9 t^2 \]
\[ \Rightarrow t \ = \ \sqrt{ \dfrac{ 3 }{ 4.9 } } \]
\[ \Rightarrow t \ = \ 0.782 \ s\]
Using the formula for speed in horizontal direction:
\[ v_x \ = \ \dfrac{ 10 }{ 0.782 } = 12.78 \ m/s \]
Calculating the magnitude of the speed:
\[ |v| \ = \ \sqrt{ v_x^2 \ + \ v_y^2 } \]
\[ \Rightarrow |v| \ = \ \sqrt{ ( 12.78 )^2 \ + \ ( 7.668 )^2 } \]
\[ \Rightarrow |v| \ = \ 14.9 \ m/s \]
Calculating the direction of the speed:
\[ \theta \ = \ tan^{-1} \bigg ( \dfrac{ v_y }{ v_x } \bigg ) \]
\[ \theta \ = \ 36.9^{ \circ } \]
Numerical Result
\[ v \ = \ 14.9 \ m/s \text{ at } \theta = 36.9^{ \circ } \text{ from ground } \]
Example
A man makes a leap $ 2.0 \ m $ long and $ 0.5 \ m $ high. What is the speed of the man just as he leaves the ground?
Using the 3rd equation of motion in the vertical direction:
\[ v_{ fy }^2 \ = \ v_{ iy }^2 + 2 a S \]
\[ \Rightarrow v_{ iy } \ = \ \sqrt{ -2 a S – v_{ fy }^2 } \]
\[ \Rightarrow v_{ iy } \ = \ \sqrt{ -2 ( -9.8 ) ( 0.5 ) – 0 } \ = \ 9.8 \ m/s \]
Using second equation of motion:
\[ S = v_{iy} t + \dfrac{ 1 }{ 2 } a t^2 \]
\[ 0.5 \ = \ ( 0 ) t + \dfrac{ 1 }{ 2 } (9.8) t^2 \]
\[ \Rightarrow t \ = \ \sqrt{ \dfrac{ 0.5 }{ 4.9 } } \ = \ 0.32 \ s \]
Using the formula for speed in horizontal direction:
\[ v_x \ = \ \dfrac{ 2 }{ 0.32 } = 6.25 \ m/s \]
Calculating the magnitude of the speed:
\[ |v| \ = \ \sqrt{ v_x^2 \ + \ v_y^2 } \ = \ \sqrt{ ( 6.25 )^2 \ + \ ( 9.8 )^2 } \ = \ 11.62 \ m/s \]
Calculating the direction of the speed:
\[ \theta \ = \ tan^{-1} \bigg ( \dfrac{ v_y }{ v_x } \bigg ) \ = \ tan^{-1} \bigg ( \dfrac{ 9.8 }{ 6.25 } \bigg ) \ = \ 57.47^{ \circ } \]