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
Step-by-step 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
![T=\sqrt{(2 h)/(g)}](https://img.qammunity.org/2020/formulas/physics/middle-school/mvdpaf8ou52jprs4gj15siwb59yrsfz97p.png)
Putting the values, we get,
![T=\sqrt{(2 * 1.19)/(9.8)}=0.49 \mathrm{sec}](https://img.qammunity.org/2020/formulas/physics/middle-school/ifhal1imit289crzy070zsyuuxnv26dprw.png)
Thereby,
![\text {distance}=\text {velocity } * \text {time}=5 * 0.49=2.45 \mathrm{m}](https://img.qammunity.org/2020/formulas/physics/middle-school/umzxmbeg9ltp00s9akzwjtcqmcz6fmplqx.png)