An egg is thrown horizontally off a roof, which is 60 meters high, with an initial velocity of 6.5 m/s. How long does it take to hit the ground, and how far does it go in the x direction?

Question

William Audy · Accepted Answer

It will take #3.5s# to reach the ground.

It will go #22.75m# in the horizontal direction.

The egg will come down because of gravity but due to its imparted horizontal velocity,it will move forward from the bottom of the roof! So,if it takes time #t# to reach the ground,considering vertical motion,we can use #h=1/2 at^2# where, #h# is the height #a=g# i.e acceleration due to gravity. Given, #h=60m# and taking #g=9.81 ms^-2# we get, #t=sqrt((2h)/g)=sqrt((2×60)/9.81)=3.5s# So,in this time if it moves a distance of #x# in the horizontal direction due to constant velocity #v=6.5 ms^-1# Then,we find #x=vt=6.5×3.5=22.75m#

Emily Abbate · Answer

Egg is thrown with horizontal velocity. Therefore, initial vertical velocity of egg #=0#. Time taken by egg to fall down under gravity and hit the ground can be found by the kinematic expression #h=ut+1/2g t^2# When we enter the given values, we #60=0xxt+1/2(9.81) t^2#
#=> t=sqrt((60xx2)/9.81)#
#=> t=3.5\ s# Distance traveled in #x# direction#=v_ht=6.5xx3.5=22.7\ m#

Amelia Adams · Answer

undefined To find the time it takes to hit the ground:
$$t = \frac{2h}{g}$$
where $h$ is the initial height (60 meters) and $g$ is the acceleration due to gravity (approximately $9.8 \, \text{m/s}^2$).

To find the distance traveled in the x-direction:
$$d = v_x \cdot t$$
where $v_x$ is the initial horizontal velocity (6.5 m/s) and $t$ is the time calculated in the first step.