Question

A tennis ball is thrown vertically upward with velocity of 49 m/s . Calculate 1) the time taken by the ball to reach the max. height . 2) max. height reached by the ball. 3) time taken by the ball to hit the ground after it starts its downward motion from the max. height attained 4) total time taken by it to return to the ground

Answer 1

I do found the 1st one ;

Time of ascent = \(\large{\frac{u sin \theta}{g}}\) = 49 / 9.8 = 490 / 98 = 5 seconds

m i right?

Answer 2

or we can do like this :

v = u -gt
0 = 49 - (9.8) T
-49/- 9.8 = T
5 seconds = Time of ascent

Answer 3

That looks right to me.

Answer 4

ok now coming to the second part:

H = \(\large{\frac{u^2 sin^2 \theta}{2g}}\)

49^2 / 2*9.8 = 49 * 49 / 19.6 = 122.5 m

Answer 5

right?

Answer 6

Looks great.

Answer 7

i am stuck at this point now :time of descent

Answer 8

Since acceleration is the same, it must be the same amount of time it took to reach the max height. Thus, the descent time is 5 seconds.

Answer 9

well right but according to the formula :
tiem fo descent = u sin theta / g

Answer 10

but when : the ball comes down the final velocity will become initial velocity there

Answer 11

right...

Answer 12

so this is ..

Answer 13

we have : v1 = 0 m/s , u1 = 49 m/s

v1 = u2 = 0 m/s
and u1 = v2 = 49 m/s

Answer 14

but in the formula :
T = u sin theta / g

we have u not v that is initial velocity 2 : u2

but u2 = 0
*u2 does not mean u^2 it means u_2

hence T = 0

Answer 15

where m i wrong? kg?

Answer 16

In this equation, \(u\) is used as the initial velocity? Or the velocity at ground?

Answer 17

I think u is used as the initial velocity wait lemme confirm it

Answer 18

i know the deriviation of that formula though .. may be that help? :

\[\large{s=ut -\frac{1}{2}gt^2}\]
\[\large{0 = usin\theta t-\frac{1}{2}gt^2}\]
\[\large{t = 0 , t = \frac{2usin\theta}{g}}\]

Answer 19

The first equation is not correct if you are starting at a non-zero height. The full equation is\[\large{s=s_0+ut -\frac{1}{2}gt^2}\]