Answered

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A body is projected horizontally from the top of a tower with a velocity of 20m/s it landed on a grand level a
t a horizontal distance of 60m from the foot of the tower. Calculate the height of the tower (g=10m/s²)

Sagot :

AshrieIIeidn

[tex]\sf\pink{.　 . • ☆ . ° .• °:. *₊ ° . ☆.　 . • ☆ . ° .• °:. *₊ ° . ☆}[/tex]

[tex]\pink{\mathbb{\huge{꧁ᬊᬁ~ANSWER~ᬊ᭄꧂}}}[/tex]

Step-by-step solution:

1) Given:

[tex]\sf\pink{ᯓ★}[/tex] Initial horizontal velocity (v₀): 20 m/s

[tex]\sf\pink{ᯓ★}[/tex] Horizontal distance traveled (d): 60 m

[tex]\sf\pink{ᯓ★}[/tex] Acceleration due to gravity (g): 10 m/s²

2) Horizontal motion:

[tex]\sf\pink{ᯓ★}[/tex] Horizontal velocity remains constant: vx = v₀ = 20 m/s

[tex]\sf\pink{ᯓ★}[/tex] Horizontal distance: d = v₀ • t

3) Vertical motion:

[tex]\sf\pink{ᯓ★}[/tex] Initial vertical velocity (v₀y): 0 m/s

[tex]\sf\pink{ᯓ★}[/tex] Vertical acceleration: ay = -g = -10 m/s²

[tex]\sf\pink{ᯓ★}[/tex] Vertical displacement (height): h = [tex]\large{\frac{1}{2}}[/tex] • ay • t²

4) Equating horizontal distances:

[tex]\sf\pink{ᯓ★}[/tex] d = v₀ • t

[tex]\sf\pink{ᯓ★}[/tex] 60 = 20 • t

[tex]\sf\pink{ᯓ★}[/tex] t = [tex]\large{\frac{60}{20}}[/tex] = 3 s

5) Calculating height:

[tex]\sf\pink{ᯓ★}[/tex] h = [tex]\large{\frac{1}{2}}[/tex] • (-10) • 3²

[tex]\sf\pink{ᯓ★}[/tex] h = [tex]\large{\frac{1}2{}}[/tex] • 90

[tex]\sf\pink{ᯓ★}[/tex] h = 45 m

[tex]\sf\pink{╴╴╴╴╴⊹ꮺ˚ ╴╴╴╴╴⊹˚ ╴╴╴╴˚ೃ ╴╴}[/tex]

So in summary, with the given initial velocity of 20 m/s and the projectile landing 60 m horizontally from the base, the height of the tower must be [tex]\blue{\underline{\sf\pink{45 ~m}}}[/tex].

[tex]\bold{\small\pink{⋆˚࿔~ ashrieIIe~˚⋆}}[/tex] [tex]\pink{\heartsuit}[/tex]

[tex]\sf\pink{.　 . • ☆ . ° .• °:. *₊ ° . ☆.　 . • ☆ . ° .• °:. *₊ ° . ☆}[/tex]

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