Asked by Chan
A projectivle is launched with an initial speed of 40m/s at an angle of 35 degree above the horizontal. The projectitile lands on a hillside 4s later. negelct air friction.
a.What is the straight-line distance form where the prjectile was launched to where it hits its target?note that the hill may slope up or down from the launch point. Thank you
a.What is the straight-line distance form where the prjectile was launched to where it hits its target?note that the hill may slope up or down from the launch point. Thank you
Answers
Answered by
MathMate
Consider horizontal and vertical motions separately.
Horizontal:
velocity, v0 = 40*cos(35°) m/s
Time, t = 4 s.
Horizontal distance, H = v0*t m.
Vertical:
initial velocity, v0 = 40 sin(35°) m/s
acceleration, a = -g m/s/s/
Time, t = 4 s.
vertical distance, V = v0*t+(1/2)at²
Straignt line distance
=√(H²+V²)
Check my thinking and calculations.
Horizontal:
velocity, v0 = 40*cos(35°) m/s
Time, t = 4 s.
Horizontal distance, H = v0*t m.
Vertical:
initial velocity, v0 = 40 sin(35°) m/s
acceleration, a = -g m/s/s/
Time, t = 4 s.
vertical distance, V = v0*t+(1/2)at²
Straignt line distance
=√(H²+V²)
Check my thinking and calculations.
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