Answer:
[tex]v_i=189.315ft/s[/tex]
Explanation:
We are going to use the accelerated motion formula [tex]v_f^2=v_i^2+2ad[/tex] on the vertical component taking the upwards direction as positive.
Since at maximum height the speed will be [tex]v_f=0m/s[/tex], the formula can be written as [tex]v_i=\sqrt{-2ad}[/tex], which for our values is:
[tex]v_i=\sqrt{-2(-32ft/s^2)(560ft)}=189.314553059ft/s=189.315ft/s[/tex]
Written to three decimal places.
The initial velocity at which the object must be thrown upwards is; u = 189.31 ft/s
To solve this, we will use Newton's third equation of motion which is;
v² = u² + 2as
Where;
v is final velocity
u is initial velocity
a is acceleration
s is distance
In projectiles, the speed at the highest point which is the top of the monument here is usually 0.
Thus; v = 0 ft/s
We are given;
a = -32 ft/s
a = -32 ft/ss = 560 ft
Thus, Plugging in the relevant values into the equation gives;
0² = u² + 2(-32 × 560)
0 = u² - 35840
u² = 35840
u = √35840
u = 189.31 ft/s
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