Lavanya throws a ball upwards, from a rooftop, which is 20 m above from ground. It will reach a maximum height
and then fall back to the ground. The height of the ball from the ground at time is h, which is given by h=-4t+16+ + 20.
(i) What is the height reached by the ball after 1 second?
(a) 64 m (C) 32 m (b) 128 m (d) 20 m
(ii) What is the maximum height reached by the ball?
(a) 54 m (c) 36 m (b) 44 m (d) 18 m
(iii) How long will the ball take to hit the ground?
(a) 4 seconds (c) 5 seconds (b) 3 seconds (d) 6 seconds
(iv) What are the two possible times to reach the ball at the same height of 32 m?
(a) I and 3 seconds (c) I and 2 seconds
(v) Where is the ball after 5 seconds
(a) at the ground (c) at highest point (b) 1 and 4 seconds (d) 1 and 5 seconds (b) rebounds (d) fall back
Answers
Given : Lavanya throws a ball upwards, from a rooftop, which is 20 m above from ground. It will reach a maximum height
and then fall back to the ground
h = -4t² + 16t + 20
To Find :height reached by the ball after 1 second
maximum height reached by the ball
How long will the ball take to hit the ground?
What are the two possible times to reach the ball at the same height of 32 m?
Where is the ball after 5 seconds
Solution:
h = -4t² + 16t + 20
t = 1
=> h = -4 + 16 + 20
=> h = 32
height reached by the ball after 1 second = 32 m
h = -4t² + 16t + 20
dh/dt = -8t + 16
dh/dt = 0
=> t = 2
d²h/dt² = -8 < 0 hence max height at t = 2
h = -4(2)² + 16*2 + 20
=> h = 36
Max height = 36 m
height at ground = 0
-4t² + 16t + 20 = 0
=> t² - 4t - 5 = 0
=> (t - 5)(t + 1) = 0
=> t = 5 , - 1
Hence ball take 5 secs to hit the ground
-4t² + 16t + 20 = 32
=> t² -4t - 5= - 8
=> t² - 4t + 3 = 0
=> (t 1)(t - 3) = 0
1 and 3 sec
ball after 5 seconds at the ground
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