An arrow is fired at an angle of 53 with a velocity of 15 m/s.
what is its range?
explain step by step please!!
2 answers
and...the time is 3.06s
if it is fired from the ground, and hits the ground, then
distance=vhorizontal*timeinair
but vhorzontal is 15cos53 or 9.03m/s
time in the air is given, I think as 3.06 seconds.
Now checking the time in air given. the vertical equation is given as
vfinal=Vinitial-g *time
and vfinal is= - vinitial, so
time= 2Vinitialvertical/g
= 2*15*sin53/9.8=2.44 seconds. Hmmm. I am not certain what then you mean by time is 3.06 seconds
distance=vhorizontal*timeinair
but vhorzontal is 15cos53 or 9.03m/s
time in the air is given, I think as 3.06 seconds.
Now checking the time in air given. the vertical equation is given as
vfinal=Vinitial-g *time
and vfinal is= - vinitial, so
time= 2Vinitialvertical/g
= 2*15*sin53/9.8=2.44 seconds. Hmmm. I am not certain what then you mean by time is 3.06 seconds
4 answers