Question

A ball is launched at an angle of 33.4 degrees up from the horizontal, with a muzzle velocity of 6.5 meters per second, from a launch point which is 1 meters above the floor. How far horizontally (in meters) from the launcher will it land on the floor? Use 9.82 meters per second for “g”.

Answer 1

Initial velocity of ball=6.5m/s

Angle of projectile =33.4°

There will be two components of velocity, one along horizontal and one along vertical.

Velocity along horizontal component will be responsible for ball to travel along horizontal path.

Horizontal component of velocity =vcosx where x is angle of projectile=6.5*cos(33.4°)

V horizontal =6.5*0.835=5.43m/s

Distance traveled in horizontal =velocity in horizontal *time

On simplifying the formula of velocity and time we get the following equation

Range=V²/2g{1+(1+2gy°/V²sin²x)¹/²}sin2x where

V=initial velocity

Y°=initial highet from the ground

g=acceleration due to gravity

X=angle of protection

Put the values in above flrmula we get

Range=6.5²/2*9.8{1+(1+2*1*9.8/6.5²*0.30)¹/²}0.92

Range=42.25/19.6{1+(1+19.6/12.675)¹/²}0.92

Range=2.168{1+1.56}0.92

Range=2.168*2.56*0.92

Range=5.055m or this is the distance on floor at which the ball will fall.

This is the simplest way to solve.i have not mentioned more calculations to avoid any kind of confusion.