Question

If a ball is thrown straight upwards at a speed of 11 m/s from balcony, 4 m above the ground, how much time would it take to strike the ground at the base of a balcony.

plz plz answer it fast

Answer 1

First of all the given is
u = 11m/s

v = 0m/s. (as it reach to the maximum distance it velocity become 0)

a = -g. ( as moving opposite to the direction of accelaration due to gravity)

distance of balcony from the ground = 4m

so first we calculate how much distance did it cover when thrown vertically upward.

v^2 - u^2 = 2aS
0^2 - 11^2 = -2gS
-121 = -2*10*S
20*S = 121
S = 121/20
S = 6.05m
S = 6m (approx)

now the total distance ball have to covered should be
4 + 6 = 10m

now we can calculate the time
for this we have
S = 10
a = g ( moving downward)
u = 0 (starting from the highest point where the velocity is 0)

So,

S = ut + 1/2 at^2
10 = (0)t + 1/2 (10) t^2
10 = 5t^2
t^2 = 2
t = √(2) s
t = 1.414 s

this is the answer

Answer 2

2.5 sec

Explanation:

ball is thrown upward then g=-10m/s^2

and final velocity=0

now

v=u+at

0=11+(-10t)

t=1.1 (approx)

for distance

2as=v^2 - u^2

2×-10×s=0^2-11^2

s=6.05 (take it 6 approx)

So total disance from ground to final position of ball is 6+4=10 m(4 is height of balcony from ground)

now

When object reaches at final topmost point and starts falling downward then it's initial velocy becomes 0 and g=10 m/s^2

S=ut+1/2at^2

10m=0×t+1/2×10×t^2

2=t^2

t=1.4 sec(approx)

so total time =1.4+1.1=2.5 sec