A balloon starts rising from ground with an acceleration of 1.25 m / s ∧ 2 1.25\mathrm{m} / \mathrm{s}^{\wedge}2 1.25 m / s ∧ 2
Solution.
An equation of motion of a balloon:
h = a t 1 2 2 . h = \frac {a t _ {1} ^ {2}}{2}. h = 2 a t 1 2 .
The speed of the balloon at the height h h h
v = a t 1 . v = a t _ {1}. v = a t 1 .
A body has the same speed v = a t 1 v = at_{1} v = a t 1
An equation of motion of a body:
0 = h + v t 2 − g t 2 2 2 ; 0 = h + v t _ {2} - \frac {g t _ {2} ^ {2}}{2}; 0 = h + v t 2 − 2 g t 2 2 ; 0 = a t 1 2 2 + a t 1 t 2 − g t 2 2 2 ; 0 = \frac {a t _ {1} ^ {2}}{2} + a t _ {1} t _ {2} - \frac {g t _ {2} ^ {2}}{2}; 0 = 2 a t 1 2 + a t 1 t 2 − 2 g t 2 2 ; 0 = a t 1 2 g + 2 a t 1 g t 2 − t 2 2 . 0 = \frac {a t _ {1} ^ {2}}{g} + \frac {2 a t _ {1}}{g} t _ {2} - t _ {2} ^ {2}. 0 = g a t 1 2 + g 2 a t 1 t 2 − t 2 2 . t 2 t_2 t 2
There is a quadratic equation, where t 2 t_2 t 2
t 2 2 − 2 a t 1 g t 2 − a t 1 2 g = 0 ; t _ {2} ^ {2} - \frac {2 a t _ {1}}{g} t _ {2} - \frac {a t _ {1} ^ {2}}{g} = 0; t 2 2 − g 2 a t 1 t 2 − g a t 1 2 = 0 ;
A discriminant is:
Δ = ( 2 a t 1 g ) 2 − 4 ⋅ ( − a t 1 2 g ) = 4 a 2 t 1 2 g 2 + 4 a t 1 2 g = 4 a t 1 2 g ( a g + 1 ) . \Delta = \left(\frac {2 a t _ {1}}{g}\right) ^ {2} - 4 \cdot \left(- \frac {a t _ {1} ^ {2}}{g}\right) = \frac {4 a ^ {2} t _ {1} ^ {2}}{g ^ {2}} + \frac {4 a t _ {1} ^ {2}}{g} = \frac {4 a t _ {1} ^ {2}}{g} \left(\frac {a}{g} + 1\right). Δ = ( g 2 a t 1 ) 2 − 4 ⋅ ( − g a t 1 2 ) = g 2 4 a 2 t 1 2 + g 4 a t 1 2 = g 4 a t 1 2 ( g a + 1 ) . t 2 ′ = 2 a t 1 g + Δ 2 = 2 a t 1 g + 4 a t 1 2 g ( a g + 1 ) 2 = 2 a t 1 g + 2 t 1 a g ( a g + 1 ) 2 = a t 1 g + t 1 a g ( a g + 1 ) . t _ {2} ^ {\prime} = \frac {\frac {2 a t _ {1}}{g} + \sqrt {\Delta}}{2} = \frac {\frac {2 a t _ {1}}{g} + \sqrt {\frac {4 a t _ {1} ^ {2}}{g} \left(\frac {a}{g} + 1\right)}}{2} = \frac {\frac {2 a t _ {1}}{g} + 2 t _ {1} \sqrt {\frac {a}{g} \left(\frac {a}{g} + 1\right)}}{2} = \frac {a t _ {1}}{g} + t _ {1} \sqrt {\frac {a}{g} \left(\frac {a}{g} + 1\right)}. t 2 ′ = 2 g 2 a t 1 + Δ = 2 g 2 a t 1 + g 4 a t 1 2 ( g a + 1 ) = 2 g 2 a t 1 + 2 t 1 g a ( g a + 1 ) = g a t 1 + t 1 g a ( g a + 1 ) . t 2 ′ = 1.25 ⋅ 8 9.8 + 8 1.25 9.8 ( 1.25 9.8 + 1 ) = 4.05 ( s ) . t _ {2} ^ {\prime} = \frac {1 . 2 5 \cdot 8}{9 . 8} + 8 \sqrt {\frac {1 . 2 5}{9 . 8} \left(\frac {1 . 2 5}{9 . 8} + 1\right)} = 4. 0 5 (s). t 2 ′ = 9.8 1.25 ⋅ 8 + 8 9.8 1.25 ( 9.8 1.25 + 1 ) = 4.05 ( s ) . t 2 ′ ′ = 2 a t 1 g − Δ 2 = 2 a t 1 g − 4 a t 1 2 g ( a g + 1 ) 2 = 2 a t 1 g − 2 t 1 a g ( a g + 1 ) 2 = a t 1 g − t 1 a g ( a g + 1 ) . t _ {2} ^ {\prime \prime} = \frac {\frac {2 a t _ {1}}{g} - \sqrt {\Delta}}{2} = \frac {\frac {2 a t _ {1}}{g} - \sqrt {\frac {4 a t _ {1} ^ {2}}{g} \left(\frac {a}{g} + 1\right)}}{2} = \frac {\frac {2 a t _ {1}}{g} - 2 t _ {1} \sqrt {\frac {a}{g} \left(\frac {a}{g} + 1\right)}}{2} = \frac {a t _ {1}}{g} - t _ {1} \sqrt {\frac {a}{g} \left(\frac {a}{g} + 1\right)}. t 2 ′′ = 2 g 2 a t 1 − Δ = 2 g 2 a t 1 − g 4 a t 1 2 ( g a + 1 ) = 2 g 2 a t 1 − 2 t 1 g a ( g a + 1 ) = g a t 1 − t 1 g a ( g a + 1 ) . t 2 ′ = 1.25 ⋅ 8 9.8 − 8 1.25 9.8 ( 1.25 9.8 + 1 ) = − 2.01 ( s ) . t _ {2} ^ {\prime} = \frac {1 . 2 5 \cdot 8}{9 . 8} - 8 \sqrt {\frac {1 . 2 5}{9 . 8} \left(\frac {1 . 2 5}{9 . 8} + 1\right)} = - 2. 0 1 (s). t 2 ′ = 9.8 1.25 ⋅ 8 − 8 9.8 1.25 ( 9.8 1.25 + 1 ) = − 2.01 ( s ) .
The time cannot be negative, then t 2 = t 2 ′ = 4.05 s t_2 = t_2' = 4.05s t 2 = t 2 ′ = 4.05 s
**Answer:** The time taken by the body to reach the ground is t 2 = 4.05 s t_2 = 4.05s t 2 = 4.05 s
#340153 on Dec 2023