Question

A golf ball is hit with an initial velocity of 50 m/s at an angle of 45 above the horizontal. how far will the ball travel horizontally before it hits the ground?

Answer 1

u=50
A=45
g=9.8
Range of projectile R = u² sin2A/g
R= 50²/ 9.8
= 2500/9.8
= 255.10 m
if you take g=10
then,
R = 250m

Answer 2

Heya

Horizontal Component of the Ball's velocity =
$50 \cos( {45}^{o} )$
First Let's Answer "HOW LONG"

Vertical Component =
$50 \sin( {45}^{o} )$

Gravity = ( -g )

=> t = 2( 0 - 50 sin( 45° ) )/(-g) = 10 sin( 45° )

Now, HOW FAR :

$vt = 50 \cos( {45}^{o} ) \times 10 \sin( {45}^{o} ) = 250m$

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Or,

HORIZONTAL Range :

$\frac{ v_{0}^{2} \sin(2\theta) }{g}$

$= \frac{ {50}^{2} \sin( {90}^{o} ) }{g}$
$= 250m$


∆ Note : We assume 'g' as 10 m/s^2