A projectile is fired with an inital velocity of 300 feet per second at an angle of 70 with the horizontal. The parametric equations for the path of the projectile are x = (300cos70)t, y = (300sin 70) t - 16t^2.
Everything is in degrees btw.
a. what is the maximum altitude(nearest foot) and the time (nearest second) the projectile reaches its maximum altitude.
b. what is the total flight time (nearest second) and the horizontal distance(nearest foot) traveled by the projectile.
I have no idea how to setup or do this please help.
The equation for the VELOCITY of the projectile is 300 sin 70 - 32t (according to the equation you have for accelration of gravity = 32 ft/sec^2) When this velocity in the 'y' direction eqauls 'zero' it has reached it's apex
so 0 = 300 sin70 - 32t results in t = 8.8096 seconds = 9 seconds rounded
The path equation for y (given in the question) is y = 300sin70 t - 16t^2 Substitute 't'
300 sin70 (8.8096) - 16 (8.8096)^2 = 1241.75 ft = 1242 ft rounded
WHen the POSITION of the projectile = 0 it has hit the ground (right?) ...when y=0
y= 300 sin70 t - 16t^2 = 0 SOlve for 't'
281.907 t - 16t^2 =0
281.907 t = 16t^2
281.907 = 16t
t = 17.61 When the projectile hits the ground....now you know how long it was flying....substitute this value of 't' into the x equation to find out how far downrange it went.......
