Question

When an object is projected into the air, it follows the path of a parabola. The equation always has the

same form, but the numbers change based on the data for the projectile:
If a cannonball is launched from a height of 29.4 m above the ground with an initial velocity of 24.5 m/s,
then the equation that models its path would be h(t) =
-4.9+7 + 24.5t + 29.4. This graph shows its
path: What is the height of the cannonball before it is launched,
at t=0? Remember to include units

Answer 1

29.4 m

Explanation:

From the graph

• You can clearly infer from the graph that the initial height of the cannon ball, before it is launched is 29.4 m
• Just find t = 0, and see where the line of the graph intersects the y-axis, which represents height of the cannon ball w.r.t time.

From the equation

• An even simpler method is substitute t = 0 into the equation for the motion of the cannon ball
• h(0) = -4.9(0)² + 24.5(0) + 29.4
• h(0) = 29.4 m