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

hèllo, sorry i really need some more points so if you need this qustion answered than retype it

dec 23, 2016 - a load of bricks weighing 680 n a distance of 10 m across a parking lot. if if the worker pushes the cart with a constant force of 209 n, what work= force x distance = 220 n x 10 m = 2200 n.m = 2200 j i don't know where the person got it from, but the initial part, work= force x distance me.