Question

In springboard diving, the diver strides out to the end of the board, takes a jump onto its end, and uses the resultant spring-like nature of the board to help propel him into the air. Assume that the diver’s motion is essentially vertical. He leaves the board, which is 3.0 m above the water, with a speed of 6.3 m/s. How long is the diver in the air, from the moment he leaves the board until

Answer 1

Answer:

The diver is in the air for $1.65s$.

Explanation:

Hi

• Known data $v_{i}=6.3m/s, y_{i}=3.0m$ and $g=9.8m/s^{2}$.

1. We need to find the time when the diver reaches the highest point, so we use the following equation $v_{f} =v_{i}-gt$ with $v_{f}=0ms$ so $t_{1}=\frac{v_{i}-v_{f} }{g}=\frac{6.3m/s}{9.8m/s^{2} }=0.64 s$.
2. Then we need to find the highest point, so we use $y=v_{i}t-\frac{1}{2} gt^{2}=(6.3m/s)(0.64s)-\frac{1}{2} (9.8m/s^{2})(0.64s)^{2}=2.03m$, this is above the springboard so the highest point is $y_{max}=5.03m$.

Finally, we need to find the time in free fall, so we use $y_{f}=y_{i}+v_{i}t-\frac{1}{2}gt^{2}$, at this stage $v_{i}=0m/s, y_{i}=5.03m$ and $y_{f}=0m$, therefore we have $0m=5.03-\frac{1}{2}(9.8m/s^{2})t^{2}$, and solving for $t_{2}=\sqrt{\frac{5.03m}{4.9m/s^{2}}} =\sqrt{1.02s^{2}}=1.01s$.

Last steep is to sum $t_{1}$ and $t_{2}$, so $t_{T}=t_{1}+t_{2}=0.64s+1.01s=1.65s$.