We should use the equation for the moment of flight, but we don"t have actually v0. We can discover v0 from the expression for range as:

v0=Rgsin(2θ)=(220)(9.8)sin<(2)(45°)>

v0 =46.43 m/s

An arrowhead is shot in ~ an edge of θ = 45° over the horizontal. The arrow hits a tree a horizontal distance D = 220m away, at the same height over the ground as it was shot. The elevation from i beg your pardon the arrowhead is shooting is 6 m. Use g=9.8m/s2 because that the size of the acceleration due to gravity.

You are watching: An arrow is shot at an angle of θ=45∘ above the horizontal

**Part A**

Find *t*a, the moment that the arrow spends in the air.

**Part B**

How lengthy after the arrowhead was shot need to the apple it is in dropped, in order because that the arrow to pierce the apple as the arrow hits the tree?

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What scientific ide do you require to know in bespeak to resolve this problem?

Our tutors have indicated the to deal with this difficulty you will require to apply the symmetry Launch concept. You can view video clip lessons to find out Symmetrical Launch. Or if you need much more Symmetrical start practice, you can additionally practice symmetry Launch exercise problems.

What professor is this problem relevant for?

Based on ours data, we think this problem is relevant for Professor Blake's class at MC MARICOPA.

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