An arrow is shot at 33.0 above the horizontal. Its velocity is 48 m/s, and it hits the target. What is the maximum height the arrow will attain? The target is at the height from which the arrow was shot. How far away is it?

1 Answer
Shwetank Mauria
Apr 23, 2018

Maximum height arrow will attain is #69.73# meters and target is #214.77# meters away.

Explanation:

The velocity of arrow is #48m.s# at #33^@# above horizontal,

hence while horizontal component is #48cos33^@=48xx0.8387=40.2576m/s#

and vertical component is #48sin33^@=48xx0.5446=26.1408m/s#

https://fundopma.weebly.com/projectile-motion.html

At #26.1408m/s# as acceleration due to gravity is #9.8m/s^2#, it will become #0# in #26.1408/9.8=2.6674# seconds

and hence arrow will reach a height of #2.6674xx26.1408~=69.73# meters

As the curve is symmetric it will reach the target, which is at the same height as the point from which arrow was shot after #2.6674xx2# seconds and as horizontal component of velocity is #40.2576m/s#,

target is #2.6674xx2xx40.2576~=214.77# meters.

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