Proj Motion EX 11

Helena Dedic

A projectile fired from the ground has a velocity of 24i -8j m/s at a height of 9.8 m. Find the initial velocity and the maximum height.

Solution:

We know that $\mathrm{\Delta }x=1.6m$ and $\mathrm{\Delta }y=-1.1m$.

We can find the initial velocity in the y-direction using equation 4:

$v_y^2-v_{yo}^2=-19.6m/s^2\mathrm{\Delta }y$

$v_{yo}^2=v_y^2+19.6m/s^2\mathrm{\Delta }y$

$v_{yo}^2=(-8m/s{)}^2+19.6m/s^{2}9.8m=256.08m^2/s^2$

$v_{yo}=16m/s$

Knowing v_{yo}, we can find the maximum height, H, by setting $v_y=0$ in equation 4:

$v_y^2-v_{yo}^2=-19.6m/s^2\mathrm{\Delta }y$ 

$0-v_{yo}^2=-19.6m/s^{2}H$ 

H = 13 m 

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