Asked by Sylvester
An arrow is horizontally projected from the top of a tower with an initial velocity of 15m/s.if it reaches the ground 7seconds later,calculate the height of the tower (let g=10m/s0).
Answers
Answered by
Enrique
x[f]=at^2/2+v[i]t+x[i]
Brackets dennote a subscript.
You're given x[f], a, t,and v[i]. You can just plug in their values into the formula and solve for x[i]. The horizontal velocity is irrelevant, the height of the tower is the initial position of the arrow, and final position is gonna be when it reaches the ground,or 0.
Brackets dennote a subscript.
You're given x[f], a, t,and v[i]. You can just plug in their values into the formula and solve for x[i]. The horizontal velocity is irrelevant, the height of the tower is the initial position of the arrow, and final position is gonna be when it reaches the ground,or 0.
Answered by
Henry
h = 0.5g*t^2.
Answered by
Victor
245m
Answered by
EMMANUEL
S=1/2*g*t*t, h=1/2*10*7*7 h=5*49=245m
Answered by
Ify
H=1/2gt²
H=1/2×10×7²
H=1/2×10×49
H=1/2×490
H=245m
H=1/2×10×7²
H=1/2×10×49
H=1/2×490
H=245m
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