A plane diving with constant speed at an angle of 53.0 with the vertical,release a projectile at an altitude of 730m. The projectile hits the ground hits the ground 5.00s after release.How far does the projectile travel horizontally during its flight?

What are the horizontal and vertical components of its velocity just before the striking the ground?

1 Answer
Nov 4, 2015

#(a). 808.6"m"#

#(b). 161.6"m/s"#

#(c). 121.9"m/s"#

Explanation:

#(a).#

MF
We can start by finding the velocity #u# just as it is released.

#s=ut+1/2"g"t^2#

Considering the vertical component#rArr#

#730=ucos53xx5+1/2xx9.8xx5^2#

#:.3u=730-122.5#

#u=202.5"m/s"#

The horizontal component of velocity #=usin53#.

We get the range from:

#s=vxxt=usin53xx5#

#:.s=202.5xx0.798xx5#

#s=808.6"m"#

You should now be able to see how I worked out (b) and (c).