A ball is thrown straight up in the air at an initial speed of 30 m/s. At the same time the ball is thrown, a person standing 70 m away begins to run toward the spot where the ball will land. How fast will the person have to run to catch the ball just before it hits the gro

11.44 m/s

Explanation:

We are given;

Initial Speed of ball;v_o = 30 m/s

Distance from person to ball;d = 70 m

Now, from projectile motion, the ball will hit the ground in time given by the formula;

t = (2v_o)/g

Wherr;

t is time taken

v_o is initial velocity

g is acceleration due to gravity

Plugging in the values with g = 9.8 m/s² as acceleration due to gravity, we have;

t = 2(30)/9.8

t = 6.12s

Now, the person will need to run 70m in 6.12s

We know that velocity/speed = distance/time

Thus;

Speed = 70/6.12

Speed = 11.44m/s

no sign of dia of fuel cell

  distance on the y-axis and time on the x-axis.

explanation:

the instantaneous velocity = change in position/time taken

the slope of the graph should give instantaneous velocity.

  slope of a graph = change in value of y -axis / change in values of x -axis

  comparing both the equations

    change in value of y -axis = change in position

  change in values of x -axis = time taken

so position values should be on the y axis and time values should be on the x axis.

distance on the y-axis and time on the x-axis.