Question 134025: The formula for the height, h metres, of an object propelled into the air is
h = -1/2g t squared ( only the t is squared) + vt + h
where g represents the acceleration due to gravity, which is about 9.8 m/s squared on the Earch. t seconds represents the time, v metres oer secibd reoresents the initial velocity of 34.3 m/s and is launched 2.1 m above the ground.
a) What is the maximum height, in metres, reached by the projectile?
b) How many seconds after launch does the projective reach its maximum height?
Answer by scott8148(6628)   (Show Source):
You can put this solution on YOUR website! b) the maximim value is on the axis of symmetry of the parabola
__ t=-b/2a __ t=-34.3/-9.8 __ t=3.5
a) h=-4.9t^2+34.3t+2.1 __ substituting __ h=-4.9(3.5)^2+34.3(3.5)+2.1 __ h=62.125
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