It will take 4 seconds for the slower stone to return to the ground.
The faster stone will reach a maximum height of 9 H
Further explanation
Acceleration is rate of change of velocity.
\(\large {\boxed {a = \frac{v - u}{t} } }\)
\(\large {\boxed {d = \frac{v + u}{2}~t } }\)
a = acceleration ( m/s² )
v = final velocity ( m/s )
u = initial velocity ( m/s )
t = time taken ( s )
d = distance ( m )
Let us now tackle the problem !
Given:
v₁ = 3 v₂
t₁ = 12 s
h₂ = H
Unknown:
t₂ = ?
h₁ = ?
Solution:
We can use the following formula to calculate time taken to reach the ground,
\(h = v ~ t - \frac{1}{2} ~ g ~ t^2\)
\(0 = v ~ t - \frac{1}{2} ~ g ~ t^2\)
\(v ~ t = \frac{1}{2} ~ g ~ t^2\)
\(v = \frac{1}{2} ~ g ~ t\)
\(2 ~ v = g ~ t\)
\(\large {\boxed {t = \frac{2 ~ v}{g} } }\)
Next, we will compare the time taken by these two stones to reach the ground
\(\frac{t_1}{t_2} = \frac{2 ~ v_1 / g}{2 ~ v_2 / g}\)
\(\frac{t_1}{t_2} = \frac{v_1}{v_2}\)
\(\frac{12}{t_2} = \frac{3 v_2}{v_2}\)
\(\frac{12}{t_2} = 3\)
\(t_2 = 12 \div 3\)
\(\boxed {t_2 = 4 ~ seconds}\)
We can use the following formula to find the maximum height reached by these two stones
\(v_f ~ ^2 = v ~ ^2 - 2 g h\)
\(0^2 = v ~ ^2 - 2 g h\)
\(2 g h = v ~ ^2\)
\(\large {\boxed {h = \frac{v ~ ^2}{2 g} } }\)
Next, we will compare the maximum height reached by these two stones
\(\frac{h_1}{h_2} = \frac{v_1 ~ ^2 / 2 g}{v_2 ~ ^2 / 2 g}\)
\(\frac{h_1}{h_2} = \frac{v_1 ~ ^2}{v_2 ~ ^2}\)
\(\frac{h_1}{H} = \frac{(3 v_2) ~ ^2}{v_2 ~ ^2}\)
\(\frac{h_1}{H} = \frac{9 v_2 ~ ^2}{v_2 ~ ^2}\)
\(\frac{h_1}{H} = 9\)
\(\boxed {h_1 = 9 H}\)
Learn more
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
Answer details
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
