Respuesta :
Answer:
a) 8.1 m
b) 2.306 s
Explanation:
To calculate the highest vertical distance travelled, we will Use the equations of motion,
u = initial velocity of the object = 10 m/s
v = velocity of the object at maximum height reached (the vertex) = 0 ft/s
y = highest vertical height reached = ?
y₀ = initial height of the object = 3 m
g = acceleration due to gravity = -9.8 m/s²
v² = u² + 2g(y - y₀)
0² = 10² + (2)(-9.8)(y - 3)
(y - 3) = 100/(2×9.8)
y - 3 = 5.102
y = 5.102 + 3 = 8.102 m
b) For the time taken for the object to hit the ground.
We will calculate the time taken to reach maximum height and add it to the time to reach the ground from maximum height.
Time taken to reach maximum height
u = initial velocity of the object = 10 m/s
v = velocity of the object at maximum height reached = 0 m/s
g = acceleration due to gravity = -9.8 m/s²
t₁ = ?
v = u + gt₁
0 = 10 + (-9.8)t₁
t₁ = (10/9.8)
t₁ = 1.02 s
Time taken to reach the ground from maximum height
u = initial velocity of the object at maximum height = 0 m/s
t₂ = ?
y = 8.102 m
g = acceleration due to gravity = 9.8 ft/s²
y = ut + (1/2)gt₂²
8.102 = 0 + (1/2)(9.8)t₂²
t₂² = (2×8.102)/9.8
t₂² = 1.653
t₂ = 1.286 s
T = t₁ + t₂ = 1.020 + 1.286
T = 2.306 s
Hope this Helps!!!
The maximum height reached by the object is 8.1 m and the time taken by the given object to hit the ground is 2.306 s.
From the equations of motion,
[tex]v^2 = u^2 +2g(y -y_0)[/tex]
Where,
[tex]u[/tex]= initial velocity of the object = 10 m/s
[tex]v[/tex] = velocity at maximum height reached = 0 ft/s
[tex]y[/tex] = highest vertical height = ?
[tex]y_0[/tex] = initial height of the object = 3 m
[tex]g[/tex] = gravitational acceleration = -9.8 m/s²
Put the values in the equation, solve for [tex]y[/tex]
[tex]0^2 = 10^2 + (2)(-9.8)(y - 3)[/tex]
[tex]y = 8.102\rm \ m[/tex]
b) the time taken by the object to hit the ground.
[tex]v = u + gt_1[/tex]
Put the values in the equation, solve for [tex]t_1\\[/tex]
[tex]\ 0 = 10 + (-9.8)\times t_1\\\\t_1=\dfrac { 10}{9.8}\\\\t_1= 1.02 \rm \ s[/tex]
Now, from the formula,
[tex]y = ut + (1/2)gt_2^2[/tex]
Put the values in the equation, solve for [tex]t_2[/tex]
[tex]8.102 = 0 + \dfrac 12(9.8)t_2^2[/tex]
[tex]t_2 = 1.286 \rm \ s[/tex]
So, the time to hit the ground is,
[tex]T = t_1 + t_2[/tex]
[tex]T = 1.020 + 1.286\\\\T = 2.306 \rm \ s[/tex]
Therefore, the time taken by the given object to hit the ground is 2.306 s.
To know more about velocity,
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