Question

An object is shot from the ground at 125m/s at an angle of 30o above the horizontal. How far away does the object land?

Answer 1

Answer:

R=u2sin2theta/g

R=125×125sin2(30°) /10

R= 125 × 125 ×√3/20

R= 15625√3/20

R = 781.25(√3)

R= 1353.16

Thus, range is 1352.16m

Answer 2

Answer:

The distance at which the object is going to land is 1,350 m

Explanation:

First to find the horizontal and vertical components

$v_{x}=v_{0} \cos \theta$

$v_{x}=\left(125 \frac{\mathrm{m}}{\mathrm{s}}\right) \cos \left(30^{\circ}\right)=108 \frac{\mathrm{m}}{\mathrm{s}}$

$v_{y}=\left(125 \frac{\mathrm{m}}{\mathrm{s}}\right) \sin \left(30^{\circ}\right)=62.5 \frac{\mathrm{m}}{\mathrm{s}}$

$v_{f}=v_{0}+a t$

t=6.25s,

Total time in the air is therefore "12.56" touche this value

Finally to find distance traveled by multiplying horizontal velocity and time.

$d_{x}=v_{x}(t)$

$d_{x}=108 \frac{m}{s}(12.5 s)=1,350 \mathrm{m}$

=1,350m.