Asked by Jeff
An arrow is shot from a height of 1.5 m toward a cliff of height . It is shot with a velocity of 30 m/s at an angle of 60 deg above the horizontal. It lands on the top edge of the cliff 4.0 s later. (a) What is the height of the cliff? (b) What is the maximum height reached by the arrow along its trajectory? (c) What is the arrow’s impact speed just before hitting the cliff?
Answers
Answered by
Steve
you know the height y is
y(t) = 1.5 + 30*.866t - 4.9t^2
= 1.5 + 25.98t - 4.9t^2
y(4) = 27.02
max height is y(2.65) = 35.94
vertical speed is 25.98-9.8*4 = -13.22
horizontal speed is 15.00
So, speed at impact is 20.00 m/s
y(t) = 1.5 + 30*.866t - 4.9t^2
= 1.5 + 25.98t - 4.9t^2
y(4) = 27.02
max height is y(2.65) = 35.94
vertical speed is 25.98-9.8*4 = -13.22
horizontal speed is 15.00
So, speed at impact is 20.00 m/s
Answered by
emma
this sucks no help
Answered by
bob
answer is 1
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