A ball is thrown vertically upwards with a velocity of 20ms^-1 from the top of a multistorey building. The height of the point from where the ball is thrown is 25.0m from the ground.
(a) How high will the ball rise?
(b) how long will it be before the ball hits the ground?
Answers
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Answer:
a) s = 20m
b) t = 5s
Explanation:
a)
initial velocity = 20m/s
final velocity (at a maximum height) = 0
a = -g = -10m/s^2
TO FIND : s
v^2 = u^2 + 2as
s = (v^2-u^2)/2a
s = (0-20^2)/(2* -10)
S = -400/-20 = 20m
b)
TO FIND : the total time taken right before the ball hits the ground.
t = (Time taken by the ball from A to B) + (time taken to by the ball from B to C right before hitting the ground.)
finding t₁ of A to B.
given, u=20 m/s, v = 0, s = 20, a = -10 m/s^2
to find t1, we can either use v = u+at₁ or s= ut₁ + 1/2at₁²
for s = ut₁+ 1/2 at₁², t₁ is in two terms. so this equation can be used to find time only if ut₁ or 1/2 at₁²= 0. ie, if u or a = 0.
so, we’ll use v = u+at₁
=> t₁ = (v-u)/a = (-20)/(-10) = 2 s.
finding t₂ of B to C.
here, u = 0. a = -10m/s^2. v is not given.
and s = 20+25 = 45m.
s is displacement. so it can be negative as well. is 45m negative? yes! we consider the direction at which the ball is thrown as positive, and so the direction opposite to this will be negative.
therefore, s = -45m.
again note that v is not given. and u = 0.
so we use the equation, s = ut₂+ 1/2 at₂²
=> -45 = 0(t₂) + -5t₂²= -5t₂²
=> -45/-5 = t₂²
=> 9 = 9 t₂²
=> 3 = t₂
therefore t = 2+3 = 5s
