Answer to Question #140138 in Physics for Derrick Ofori-Yeboah

Question

You launch a cannon ball at at angle of 60 degrees at 50 m/s
The horizontal component is 50*Cos 60= 25 m/s and the vertical component is 43.3 m/s found by multiplying 50 by the sine of 60 degrees
SHOW WORK- what is the total time the cannon ball is in the air?
________ sec
Find the range- where will be ball land on the ground, the target Show WORK- what formula did you use to find the answer
________meters,m

Answer 1

$$v_y=gt\\t=\frac{v_y}{g}=\frac{43.3}{9.8}=4.42\ s$$

$$R=\frac{2v_xv_y}{g}=\frac{2(25)(43.3)}{9.8}=221\ m$$