Question

A baseball is thrown with an initial velocity of 100 m/s at an angle of 30 degree above the horizontal.
How far from the throwing point will the baseball attain its original level?​

Answer 1

Answer:

500√3 meters

Explanation:

use formula of horizontal range of oblique projectile

Answer 2

Answer:

433

Explanation:

$R = \frac{ {u}^{2} \sin2 \theta }{2g}$

Given

u = 100 m/s

angle = 30 deg

g = 10 m/s2

$R = \frac{ {100}^{2} \sin(60) }{20} = 433$

Hope it helps