a) when t = 0, h = 5 meters
b) when is h = 0 (besides before the launch :)
0 = - 5 t^2 + 10 t+ 5
solve quadratic
t = -.41 and +2.41 seconds
use 2.41
c)change in height = - 5 meters
change in time =2.41 s
so
v av = -5/2.41
d)zero of course, it is stopped at the top (unless it also had horizontal speed)
I suppose you could find the vertex of the parabola
The height in metres of a projectile launched from the top of a building is given by h(t) =- 5t2 + 10t+5 where t is the time in seconds since it was launched. a) How high was the projectile at the moment of launch? b) At what time does the projectile hit the ground? c) What is the average rate of change in height from the time the object was launched until the time it hit the ground? d) What is the velocity of the projectile when it's at its maximum height. Justify your answer algebraically. [5A]
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