Question

A ball is projected horizontally above level ground from the top of a vertical cliff. the ball strikes the level ground 1 km from the base of the cliff 4 seconds after it is fired. the heigh of the cliff is:

i lost my mind help

Answer 1

`h=v_i_yt+(1)/(2)gt^2\\ \\h=(1)/(2)(9.81m/s^2)(4s)^2\\`

h = 78.48 m

Answer 2

78.48 m.

Step-by-step explanation:

• Since the ball is projected horizontally, its initial vertical velocity is zero. Therefore, we can use the following equation to find the height of the cliff:
  h = (1/2)gt^2

where h is the height of the cliff, g is the acceleration due to gravity, and t is the time it takes for the ball to hit the ground.

• We know that the ball hits the ground 1 km from the base of the cliff, so the horizontal distance traveled by the ball is 1 km. We can use the following equation to find the time t it takes for the ball to travel this distance:
  d = vt

where d is the distance traveled, v is the initial horizontal velocity, and t is the time it takes to travel the distance.

Since the ball is projected horizontally, its initial horizontal velocity is constant, and we can assume it is the same as the speed at which it hits the ground. Therefore, we can write:

• t = d/v

Substituting in the given values, we get:

• t= 1000 m / 250 m/s

• t = 4 s

Therefore, it takes the ball 4 seconds to hit the ground. Now we can use the first equation to find the height of the cliff:

• h = (1/2)gt^2

• h = (1/2)(9.81 m/s^2)(4 s)^2

• h = 78.48 m

Therefore, the height of the cliff is 78.48 m.