physicsquestion 7
posted by kelly .
A baseball is thrown at an angle of 26 degrees relative to the ground at a speed of 22.1 m/s. The ball is caught 39.2327 m from the thrower.
The acceleration due to gravity is 9.81 m/s2 .
How high is the tallest spot in the ball’s path? Answer in units of m

Vo = 22.1m/s @ 26deg.
Yo = ver. = 22.1sin26 = 9.69m/s.
hmax = (Yf^2  Yo^2) / 2g,
hmax = (0  (9.69)^2) / 19.6 = 4.79m.
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