A book is pushed with an initial horizontal velocity of 5.0 meters per second off the top of a 1.19 meter high desk. The book lands at 2.45 m.
Answer: 2.45 m
Explanation:
The book when pushed with the initial velocity of 5 m/s, the book flights down from the high desk which is 1.19 m high. The book takes the projectile motion path and hence to calculate the distance, how far it lands away from the desk can be calculated as follows,
Given that,
v = 5 m/s
h = 1.19 m
As we know that the time can be calculated as the time of flight, by
[tex]T=\sqrt{\frac{2 h}{g}}[/tex]
Putting the values, we get,
[tex]T=\sqrt{\frac{2 \times 1.19}{9.8}}=0.49 \mathrm{sec}[/tex]
Thereby,
[tex]\text {distance}=\text {velocity } \times \text {time}=5 \times 0.49=2.45 \mathrm{m}[/tex]
