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A weather balloon is launched from the ground at a spot 250 yards from a 65 foot royal palm tree. What is the minimal angle of elevation at which the balloon must take off in order to avoid hitting the treeAssume that the balloon flies in a straight line and the angle of elevation stays constant

User Liah
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6.1k points

1 Answer

12 votes

Answer:

Angle of elevation at which the balloon must take off in order to avoid hitting the tree is 5 degrees

Step-by-step explanation:

Given:

The height of the tree = 65 foot

The distance between the tree and the spot from which the balloon is launched = 250 yards

To find:

The angle of elevation at which the balloon must take off in order to avoid hitting the tree = ?

Solution:

Converting the yards to feet

1 yards = 3 feet

250 yards = 3 x 250 = 750 yards

Refer the below figure, The angle x is the angle of elevation at which the the ball must be thrown so that it does not Hit the tree

Substituting the values

User MhdSyrwan
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6.4k points
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