A small metal ball thrown vertically upwards from the top of a tower with an initial velocity of 20ms-1. If the ball took a total of 6s to reach ground level, determine the height of the tower [g = 10ms-2]

Timmynice

4 years ago

You can solve it like this
H=ut -1/2gt2
H=20*6 - 0.5*10*36
H=120-180
H=60

Timmynice

4 years ago

you can easily solve it like this with this formula

dimowocosmas

8 years ago

I love this work. More power to your elbow myschool.

Gracess

2 years ago

S=u+v then in bracket then t
_____
2

S=20
___. in bracket then 6
2
(10)6
60

Dave333

4 years ago

🤔Hmm... it's kinda long I guess

ChinonsoGlory

2 years ago

Up

buster22

1 year ago

The correct way of solving this is through second equation of motion
my school just make it long unnecessary

Henriz

4 years ago

I think the answer is 80m because from the top to the ground is actually the height of the tower. I.e the total height is 100m ( so 100m - 20m still give 80m .. Anyone who agree pls indicate.

Osayamonella

2 years ago

what of....
v=u-gt
v=20-10*6
10*6=60

Aiwansedoetionsa

3 years ago

Myschool 80m is correct. From the calculation the total distance is supposed to be 100m and not 80m while the maximum hieght attained when dispaced from the tower was 20m hence to get the height of the tower it becomes (100-20 = 80m) .. . another way you could solve this is , if the time it took the ball to go up & back to the tower was 2s the time it would takw from the top of the tower than to the ground is 4s Let relate the time with distance

H = Ut + 1/2gt^2

Coming down the initial velocity is zero

H = 1/2 gt^2

H = 1/2 x 10 x (4^2)

H = 5 x 16

H = 80m..