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.
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
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