If a projectial is fired at such an angle that the vertical componet of its velocity is 49m/s. The horizonal componet of its velocity is 61m/s, then how long will the projectia stay in the air? What distance will it travel? And what is the inital speed?
Physics - bobpursley, Sunday, January 8, 2012 at 5:26pm
Your teacher is pretty easy.
vertical component: solve for time in air.
hf=ho+vyi*t- 4.9t^2 vyi=49, hf=ho=0
He used the general equation for a body acted upon by acceleration a and initial velocity Vi
x = Xi + Vi t + (1/2) a t^2
in the up direction with Vi up = 49 an g down of 9.8 we have
height = initial height + 49 t - (1/2)(9.8) t^2
0 = 0 + 49 t - 4.9 t^2
t^2 - 10 t = 0
t = 0 of course or
t = 10 seconds in the air
The horizontal problem is just a constant speed problem
distance = speed * time
distance traveled:
distance= 61*timeinair
Initial speed= sqrt(49^2+61^1)
bobpursley can you show the equasions you are using
1 answer
1 answer