a) yes, but I wonder how you got the time to max height.
b) I have no idea what "right before...returns to ground" means. The wording is very poor. In physics, we consider gravity returning the projectile to ground starting at launch and working until it strikes the ground.
A projectile is launched from ground level with an inital velocity of 50 m/s at an angel of 60 degreess aove the horizontal. Use g=10m/s. What is the max height reached by the projectile?
I used change in y=voyt +1/2at^2
=50(2.4) + 1/2(=10) (2.5)^2
=125-31.25
=93.75m
I was just wondering if i did this correctly.
b) What is the vertical component of the final velocity right before the projectile returns to the ground?
I used vy= v sin theda
but my roblem with this is what would v equal to ?
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