Question

A diver leaves a 3-m board on a trajectory that takes her 2.2m above the board and then into the water 2.6m horizontally from the end of the board.

Part A At what speed did she leave the board? Express your answer to two significant figures and include the appropriate units. v = SubmitMy AnswersGive Up Part B At what angle did she leave the board? Express your answer using two significant figures. ? =   ?   SubmitMy AnswersGive Up

Answer 1

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. If this helps then kindly rate 5-stars

A diver leaves a 3-m board on a trajectory that takes her 2.2m above the board and then into the water a horizontal distance of 2.5m from the end of the board.

At what speed did she leave the board?
&
What was the angle?

solution:

1) you can calculate (u sin ?) vertical component of velocity by energy conservation
at start point > 0.5 mu^2 + mgh
max height point > 0.5 m[u cos ?]^2 + mgh1
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at max height point verticla speed =0, and only horizontal speed ( u cos ?) exists
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0.5 mu^2 + mgh = 0.5 m[u cos ?]^2 + mgh1
0.5u^2 [1 - cos^2 ?] = g (h1-h) = 9.8*(5.2-3)
u^2 sin^2 ? = 2*2.2*9.8 = 43.12
u sin ? = 6.57 m/s
but you have to solve for time of flight > to involve (2.9 m) Range
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origin at waterline just below the launch point
x = u cos ? * t
y - h = u sin ? * t - 0.5 gt^2
striking point will have
> y = 0, x = 2.9
2.9 = u cos ? * T
0 - 3 = u sin ? * T - 0.5 g T^2 = 6.57 T - 0.5*9.8T^2
(used above component)
3 = - 6.57 T + 4.9T^2
4.9T^2 - 6.57 T -3 =0
T = 1.70 s
T = - 0.36 (ignore negative time)
>> 2.9 = u cos ? * T
u cos ? = 2.9/1.7 = 1.7
u sin ? = 6.57
square add
u^2 = 46.1
u = 6.79 m/s
tan ? = 6.57/1.7 = 3.8647
? = 75.49 degree