A man throws a brick upward from the top of a 50 m building. The brick has an initial upward velocity of 20 m/s. a) How high above the building does the brick get before it falls? b) How much time does the brick spend going upwards? c) What is the velocity of the brick when it passes the man going downwards? d) What is the velocity of the brick when it hits the ground? e) At what time does the brick hit the ground?
Answers
Answer:
I). 20.04 meters
II). 2 sec.
III). 19.8 meters
IV). 37.05 m/s
V) 3.94 sec.
Explanation:
distance equals initial velocity multiplied by time wholly added with half multiplied by acceleration multiplied by time square.
here, distance is + x
here u= 20m/s
here a=-9.8( against gravity) or use ten
here, v = 0m/s
by,
v^2-u^2=2as
0^2-20^2=2*-9.8*s
-400=-19.6*s (cancel the minus signs)
400/19.6=s
s=20.04 meters
ii)
v=u +at
a=(v-u)/t
t=(v-u)/a
t=(0-20)/-9.8
t=-20/-9.8. (cancel the minus signs)
t=2.0r sec.
t=2s
iii)
u = 0m/s
v= x m/s
a = 9.8 m/s^2 (it is going towards the gravitational pull)
s = 20.04 meters
by,
v^2-u^2=2as
x^2-0^2=2(9.8)(20.04)
x^2=392.78
x=under root 392.78
x=19.8m/s
iv)
u=0m/s
v=x m/s
s=20.04 + 50 (height of the building and the height when thrown)
a=9.8m/s^2
v^2-u^2=2as
x^2- 0^2=2(9.8)(70.04)
x^2=1372.78
x=37.05m/s
v)
v=u +at
a=(v-u)/t
t=(v-u)/a
time when the brick was thrown upwards
t=(0-20)/-9.8
t=20/9.8s. ( cancel the minus signs)
t= 2.04 s
t=2 s........ equation 1
time when the brick falls
u=0m/s
v=37.05m/s
s=20.04 + 50 (height of the building and the height when thrown)
v=u +at
a=(v-u)/t
t=(v-u)/a
t=(37.05-20)/9.8
t=17.05/9.8 s
t=1.94...........equation 2
add equation 1 and 2
2 sec + 1.94 sec =total time
3.94 sec
approximately 4 secends
hope tis helps