Answer :
Answer:
Approximately [tex]2.55\; {\rm s}[/tex], assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex] and that air resistance on the object is negligible.
Explanation:
Assume that the air resistance on the object is negligible. The kinetic energy of the object right before the object lands will be the same as that when the object was just tossed up. As a result, the velocity of the object right before landing will be the equal in magnitude as the initial velocity, but in the opposite direction.
Initial velocity of the object: [tex]u = 12.5\; {\rm m\cdot s^{-1}}[/tex].
Final velocity of the object right before landing: [tex]v = (-12.5)\; {\rm m\cdot s^{-1}}[/tex] (same in magnitude as the initial velocity, but opposite in direction.)
Change in velocity:
[tex]\begin{aligned} \Delta v &= (v - u) \\ &= (-12.5)\; {\rm m\cdot s^{-1}} - (12.5\; {\rm m\cdot s^{-1}}) \\ &= (-25.0)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].
Also because the air resistance on the object is assumed to be negligible, the acceleration of the object will be [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex].
The time the object stays in the air is equal to the time [tex]t[/tex] required for the velocity of the object to change from [tex]u[/tex] to [tex]v[/tex]:
[tex]\begin{aligned} t &= \frac{\Delta v}{a} \\ &= \frac{(-25.0)\; {\rm m\cdot s^{-1}}}{(-9.81)\; {\rm m\cdot s^{-2}}} \\ &\approx 2.55\; {\rm s} \end{aligned}[/tex].
The object stays in air for 2.55 s.
What is acceleration due to gravity?
The acceleration obtained by a object moving under gravitational force is acceleration due to gravity. Due to having both magnitude and direction, It is a vector quantity. It is expressed by symbol g and SI unit of it is m/s. the value of g is 9.8 m/s on earth-surface at sea level acting downwards direction.
Given that,
The object is tossed in air with initial velocity (u) =12.5 m/s.
So, the object will moving upwards first and after obtaining maximum height it will fall.
During upward motion,
Initial velocity (u) = 12.5 m/s
Final velocity (v) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/[tex]s^{2}[/tex] (acting downwards)
For upward motion,
v = u - gt₁
⇒ 0 = 12.5 - 9.8×t₁
⇒ t₁ =[tex]\frac{12.5}{9.8}[/tex] = 1.275 s.
And, let maximum height obtained by the object is H.
Then,
H = [tex]\frac{u^{2} }{2g} }[/tex] = [tex]\frac{ 12.5*12.5}{2*9.8}[/tex] = 7.97 m.
Again, for downward motion,
Maximum height ( H) = 7.97 m.
Acceleration due to gravity (g) = 9.8 m/[tex]s^{2}[/tex] (acting downwards)
And, initial velocity (v) = 0 m/s
So, time taken for downward motion (t₂) = [tex]\sqrt{\frac{2H}{g} }[/tex] s = [tex]\sqrt{\frac{2*7.97}{9.8} }[/tex] s = 1.275 s.
Hence, total time of flight =( t₁ + t₂) s = ( 1.275 + 1.275) s = 2.55 s.
Learn more about acceleration due to gravity here:
https://brainly.com/question/13860566
#SPJ1
