Question

A swimmer swimming 2.30 m/s jumps out horizontally from the starting blocks and reaches the water below 1.50s later. How far from the base of the starting block does the swimmer land in the water?

Answer 1

Step-by-step explanation:

We know that distance can be calculated using the formula:

d = v0 * t + 0.5 * g * t^2

where v0= intial velocity

t=fall time

g= free fall acceleration

Now as per the question:

initial velocity v0 = 2.30 m/s

time t = 1.50 s

Free fall acceleration g = 9.8 m/s^2

d= v0 * t + 0.5 * g * t^2

distance = 2.30 * 1.50 + 0.5 * 9.8 * 1.50^2

distance = 3.45 + 14.7

d = 18.15 m

Therefore, the swimmer lands 18.15 meters from the base of the starting block.