An athlete throws a shot at an angle of 30 degrees to the horizontal at an initial speed of 43 ft/s. It leaves his hand 7 ft above the ground.
a)Where is the shot 2 seconds later?
b)How high does the shot go?

Question

An athlete throws a shot at an angle of 30 degrees to the horizontal at an initial speed of 43 ft/s. It leaves his hand 7 ft above the ground. 
a)Where is the shot 2 seconds later?
b)How high does the shot go?

anonymous · Answer

Just need some help figuring this one out.

amistre64 · Answer

its a physics application of a quadratic function

amistre64 · Answer

$$h=\frac12gt^2+V_o~sin(\theta)~t+h_o$$

amistre64 · Answer

g = gravity constant
Vo is the initial velocity
theta is the inital angle it is projected from
and ho is the initial height