An arrow is shot at 30.0 degrees above the horizontal. Its velocity is 49 m/s, and it hits the target.
a. What is the maximum height the arrow will attain?
b. The target is at the height from which the arrow was shot. How far away is it?
2007-03-07 12:55:50 · 2 answers · asked by jenergizer303 1 in Science & Mathematics ➔ Physics
The vertical component of velocity at release is 49Cos60 = 49/2 m/s.
It undergoes a uniform deceleration under the effect of gravity.
At the highest point v = 0
use v^2 = u^2 - 2gS
0 = (49/2)^2 - 2*10*S
S = (49/2)^2 / (2*10) = 30 m approx. This is the maximum height.
The time of flight = 2 * time needed to reach this maximum point.
Time to reach maximum point can be found by
v = u - gt
=> 0 = 49/2 - 10t
=>t = 49/20 s
The total time of flight is 2t = 49/10 s
In this time the horizontal distance travelled would be
Horizontal component of velocity * t (Horizontal component remains constant as there is no force in this direction).
= (49/10) * 49 * sqrt(3)/2 = 208 m
2007-03-07 16:56:00 · answer #1 · answered by FedUp 3 · 0⤊ 0⤋
25 feet for a.
b. 250 feets
2007-03-07 13:04:18 · answer #2 · answered by Joel H 4 · 0⤊ 0⤋