Question

A shot leaves the gun at a speed of 160m/s calculate thegreatest distance & height at which it would rise

Answer 1

for greatest height,
$maximum \: height = \frac{ {v}^{2} { \sin( \alpha ) }^{2} }{2g}$
$\alpha = \frac{\pi}{2}$
therefore range = 160×160/2×10
=1280 meters

for greatest distance (range),
$range = \frac{ {v}^{2} \sin(2 \alpha ) }{g}$

here,
$\alpha = \frac{\pi}{4}$
therefore range =160×160/10
=2560 meters

Answer 2

Hello buddy,

◆ Answer-
Rmax = 2612 m
Hmax = 653 m

◆ Explaination-
# Given-
u = 160 m/s

# Solution-
Formula for range is
R = u^2 sin2θ / g

For greatest distance, θ = 45°, hence
Rmax = u^2 / g
Rmax = 160^2 / 9.8
Rmax = 2612 m

Maximum height at angle θ = 45°,
Hmax = u^2 (sinθ)^2 / 2g
Hmax = (160)^2 × (1/√2)^2 / (2×9.8)
Hmax = 653 m

Therefore, greatest distance travelled can be 2612 m and for this it would raise to height of 653 m.

Hope this helps you...