Asked by K
You can shoot an arrow straight up so that it reaches the top of a 25-m-t building. (a) How far will the arrow travel of you shoot it horizontally while pulling the bow in the same way? The arrow starts 1.45 m above the ground.(b) Where do you need to put a target that is 1.45 m above the ground in order to hit it if you aim 30° above the horizontal
Answers
Answered by
Henry
Yf^2 = \Yo^2 + 2g*h = 0.
Yo^2 = -2g*h = -2*(-9.8)*25 = 490.
Yo = 22.1 m/s.
a. Yf = Yo + g*Tr = 0.
22.1 - 9.8Tr = 0.
Tr = 2.26 s. = Rise time.
Tf = Tr = 2.26 s. = Fall time.
Xo = Yo = 22.1 m/s. = Hor. velocity.
Dx = Xo*(Tr+Tf) = 22.1 * 4.52 = 100 m.
b. Range = Vo^2*sin(2A)/g.
Range = 22.1^2*sin(60)/9.8 = 43.2 m.
Yo^2 = -2g*h = -2*(-9.8)*25 = 490.
Yo = 22.1 m/s.
a. Yf = Yo + g*Tr = 0.
22.1 - 9.8Tr = 0.
Tr = 2.26 s. = Rise time.
Tf = Tr = 2.26 s. = Fall time.
Xo = Yo = 22.1 m/s. = Hor. velocity.
Dx = Xo*(Tr+Tf) = 22.1 * 4.52 = 100 m.
b. Range = Vo^2*sin(2A)/g.
Range = 22.1^2*sin(60)/9.8 = 43.2 m.
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