Question

(b) A ball of mass 400 g is thrown with an initial velocity of 30.0 m s–1 at an angle of 45.0° to the

horizontal.
Air resistance is negligible. The ball reaches a maximum height H after a time of 2.16 s.
Calculate
2. the maximum height H of the ball,

Answer 1

Use this\[y = u \sin (\theta)t - \frac{ 1 }{ 2 }g t^{2}\]

Answer 2

other way to solve is , S=0+0.5gt*2, from where zero comes.

Answer 3

Nope. You can't have zero, because initial velocity is not zero.

Answer 4

this answer is given in mark scheme, they gave two different ways to solve

Answer 5

What's the other way?

Answer 6

s=0+0.5(9.81)(2.16)*2

Answer 7

answer is either 22.88 or 22.94

Answer 8

I'd have to say its incorrect.

Answer 9

ok, thanks for your answer .

Answer 10

Nope.

Answer 11

@ali_moazzam is also correct

Answer 12

He consider the object is falling frm H height to ground

Answer 13

So its acceleration

Answer 14

U know
Time needed for an object to go frm bottom to top is just equal to time needed to go frm top to bottom

Answer 15

So if we consider the object is going frm top to bottom then the equation will b
y=ut*.5gt^2
y=0*t+.5gt^2

Answer 16

For a falling object initial velocity u=0

Answer 17

Anyway answer should b equal. If not then tell me plz!!

Answer 18

I hv 1 more another way to solve this problem. If u wanna c then response plz!!

Answer 19

no, thanks i understood , thanks !

Answer 20

We're wrong. I read wrong. The maximum height is given as \[H = \frac{ u^{2}(\sin \theta)^{2} }{ 2g }\]