An arrow is shot at a 30 degree angle with a velocity of 39m/s. How high will it go? What horizontal distance will it travel?

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Answer 1 · 2015-12-04T00:41:19

(a)

#19.4"m"#

(b)

#134.4"m"#

(a)

Considering the vertical component ot the motion:

#v^2=u^2+2as#

This becomes:

#v^2=u^2-2gh#

#:.0=(39sin30)^2-2xx9.8xxh#

#:.0=380-19.6xxh#

#h=280/19.6=19.4"m"#

(b)

The expression for range #d# on level ground is:

#d=(v^2sin(2theta))/(g)#

#:.d=(39^2xxsin(60))/(9.8)#

#d=134.4"m"#