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Answers
Explanation:
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Answer:
Answer:
Given :-
A ball thrown vertically upward with a speed of 19.6 m/s from the top of the tower and returns to the earth in 6 seconds.
To Find :-
What is the height of the tower.
Formula Used :-
\clubsuit♣ Second Equation Of Motion Formula :
\mapsto \sf\boxed{\bold{\pink{h =\: ut + \dfrac{1}{2}gt^2}}}↦
h=ut+
2
1
gt
2
where,
h = Height
u = Initial Velocity
t = Time Taken
g = Acceleration due to gravity
Solution :-
Given :
Initial Velocity (u) = 19.6 m/s
Time Taken (t) = 6 seconds
Acceleration due to gravity (g) = - 9.8 m/s²
According to the question by using the formula we get,
\longrightarrow \sf h =\: (19.6)(6) + \dfrac{1}{2} \times (- 9.8)(6)^2⟶h=(19.6)(6)+
2
1
×(−9.8)(6)
2
\longrightarrow \sf h =\: 19.6 \times 6 + \dfrac{1}{2} \times - 9.8 \times 6 \times 6⟶h=19.6×6+
2
1
×−9.8×6×6
\longrightarrow \sf h =\: 117.6 + \dfrac{1}{2} \times - 58.8 \times 6⟶h=117.6+
2
1
×−58.8×6
\longrightarrow \sf h =\: 117.6 + \dfrac{1}{2} \times - 352.8⟶h=117.6+
2
1
×−352.8
\longrightarrow \sf h =\: 117.6 + ( - 176.4)⟶h=117.6+(−176.4)
\longrightarrow \sf h =\: 117.6 - 176.4⟶h=117.6−176.4
\begin{gathered}\longrightarrow \sf h =\: - 58.8\: \: \bigg\lgroup \small\sf\bold{\pink{Height\: can't\: be\: negative\: (-\: ve)}}\bigg\rgroup\\\end{gathered}
⟶h=−58.8
⎩
⎪
⎪
⎪
⎧
Heightcan
′
tbenegative(−ve)
⎭
⎪
⎪
⎪
⎫
\longrightarrow \sf\bold{\red{h =\: 58.8\: m}}⟶h=58.8m
{\small{\bold{\underline{\therefore\: The\: height\: of\: the\: tower\: is\: 58.8\: m\: .}}}}
∴Theheightofthetoweris58.8m.