A ball is thrown at a 30 angle above the horizontal across level ground. it is released from height of 2.00 m above the ground with a speed of 24.0m/s. how far does the ball travel horizontally before it strikes the ground?

87.44m

Explanation:

Using the equation of motion

S = ut+1/2gt²...1 where

S is the distance covered by the ball

u is the velocity = 24m/s

t is the time of flight

g is the acceleration due to gravity = 9.81m/s

theta = angle of inclination = 30°

Before we can get the distance, we need to get the time of flight

t = 2usin(theta)/g

t = 2(24)sin30°/9.81

t = 48(0.5)/9.81

t = 24/9.81

t = 2.45s

Substituting t = 2.45s into equation 1 to get the distance S

S = 24(2.45)+1/2(9.81)(2.45)²

S = 58.8+29.44

S = 87.44m

does anyone know or what?