An arrow is shot at an angle of theta=45degrees above the horizontal. The arrow

Question

An arrow is shot at an angle of theta=45degrees above the horizontal. The arrow hits a tree a horizontal distance D=220m away, at the same height above the ground as it was shot. Use g=9.8m squared for the magnitude of the acceleration due to gravity.
A. FIND TA


Part B
Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree.How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?

Explanation / Answer

The ingle of projection theta = 45 degree The horizontal distance x = 220 m As the arrow hits the tree at the same height as it was released, the vertical displacement of the arrow is y=0 Let v0 be the initial felocity of the arrow The equations of motion of the arrow are: x = v0cos45 *t y = v0si45 *t - 0.5* g*t^2 220 = v0cos45 *t 0 = v0sin45 *t - 4.9*t^2 t = v0sin45 /4.9 t = (220/cos45)sin45/4.9 t = 220/4.9 t = 44.9 s The time taken by the apple to travell 6 m is 6 = 0.5*9.8*t^2 t = 1.1 s So the apple should be dropped after time 43.8 s from the time of release of the arrow

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