Asked by Lady22
A projectile is launched with an initial speed of 40m/s at an angle of 25 degrees above the horizontal. The projectile lands 3 sec. later. Neglect air friction.
What is the projectile's velocity at the highest point of it's trajectory?
What is the straight line distace from where the projectile was launched to where it hits its target? note that the hill may slope up or down from the launch point.
I'm lost, any help would be great;)
What is the projectile's velocity at the highest point of it's trajectory?
What is the straight line distace from where the projectile was launched to where it hits its target? note that the hill may slope up or down from the launch point.
I'm lost, any help would be great;)
Answers
Answered by
drwls
At the highest point, the vertical component of velocity is zero (because it has stopped rising), and the horizontal component remains the constant value that it was at launch
V = Vx = Vo cos 25 = 36.25 m/s
Distance to landing point =
Vx * 3s = 108.8 m
V = Vx = Vo cos 25 = 36.25 m/s
Distance to landing point =
Vx * 3s = 108.8 m
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