Physicos???????

Question

A projectile launched with an intial speed of 60 m/s at an angle of 30 degrees above the horizontal. The projectile lands on a hillside 4.0 sec later. Neglect air friction. (a) What is the projectile's velcotiy at the highest point of its trajectory? (b) What is the straight- line distance from where the projectile was launched to where it hits?

Ok I found time to be 6.124....

I don't get what the question is asking?????

Answer 1

I don't agree with BEN, who has found the vertical component of the speed. The horizontal component is the one that doesn't change (neglecting air resistance), and it's

60 * cos (30 deg)
which is about 51 m/s, I think.

At the highest point of the trajectory, the vertical velocity has dropped to zero, and so the velocity is just the horizontal component, as above.

(b) After 4 seconds, the horizontal distance it has travelled is
4*60*cos(30deg). That's about 205 m, I think. To find its vertical distance, use
y = ut + 0.5g(t^2)
with u = 30 (vertical component of original velocity),
t = 4 and g = -9.8

which gives about 80.

Apply Pythagoras' to these two distances and you have the straight line distance required; my guess is a bit less than 220 m.

Answer 2

The horizontal component of projectile velocity is 60*cos(30º); without air friction this does not change. At the top of the trajectory, the proctile has no vertical velocity, therefore its total velocity is the horizontal component only. The horizontal distance it travels is the horizontal component of velocity time the total time given (4.0 sec).

NOTE: The horizontal component is the COSINE of the initial angle; to verify this, imagine the initial ange is 0 (projectile shot horizontally); then the horizontal component = total initial velocity, and only cos(0) = 1; sin(0) = 0.

Answer 3

velocity at the max point = initial speed * sin30 deg
= 60m/s * 0.5
= 30 m/s

distance = 30m/s * 4s
= 120 m

Answer 4

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