Question

A projectile is thrown at an angle 30 degree with the horizontal and Kinetic Energy 'E' calculate the Energy in terms of E at maximum height?

Answer 1

at max height the velocity is always zero. right ?

Answer 2

Vertical component will be 0

Answer 3

yah !

Answer 4

Hmax = u^2 / 8g

P.E = mu^2/8

Answer 5

At hmax ..horizontal component is u cos 30 shuld we have to consider it ? @experimentX

Answer 6

yes, you still have to consider the horizontal velocity of the projectile. It's KE can't be zero at max height.

Answer 7

So at Hmax the Energy is same as the initial Energy right?

Answer 8

no. the energy at max height must be less than the energy at the start. E at max height will be equal to Ecos30

Answer 9

Hw.......?

Answer 10

At max point the body will have P.E and K.E due to horizontal component right?

Answer 11

P Hmax = u^2 / 8g

P.E = mu^2/8

Answer 12

PE will be due to height, KE will be due to its VELOCITY. the only velocity at max height is horizontal velocity. The horizontal velocity doesn't change from the moment of launch.

Answer 13

E = P.E + 1/2 m u^2 cos^2 30

Answer 14

yes

Answer 15

mu^2/8 + 3mu^2/8 = mu^2 /2

which is same as initial K.E ie 1/2 m u^2 cos^2 30 + 1/2 m u^2 sin^2 30 = mu^2/2

Answer 16

at max height...it'll have only horizontal velocity n horizontal velocity is constant throughout the motion.

Answer 17

no. your last equation was right.\[E_{total} = PE + KE\]\[E_{total} = mgh + \frac {1}{2} mv^2\] at laungh, PE = 0, so KE = E_total. At max height, PE is also amximum, so the only portion of velocity left is the horizontal\[E_{\max} = E_{total} = mgh + \frac{1}{2}m(vcos30)^2\] @akash123 is corect, the horizontal velocity doesn't change. As I already said

Answer 18

1/2 m u^2= E

Answer 19

find u then at max height K.E.= 1/2 m (u cos(theta))^2

Answer 20

What is it that you really need to know? the TOTAL energy at max height, or the KINETIC energy at max height?

Answer 21

oh..i am Silly....thxx for reminding @JFraser i need k.E

Answer 22

K.E at Top most = 3E/4...i got it...)

Answer 23

1/2 m u^2 3/4 = 3 E/4

Answer 24

thxxx....guys for helping..)

Answer 25

Hint: https://docs.google.com/viewer?a=v&q=cache:5NBhTk-Z3swJ:www.physics.ohio-state.edu/~klaus/f11-111/Solutions-Thompson.pdf+A+projectile+is+thrown+at+an+angle+30+degree+with+the+horizontal+and+Kinetic+Energy+'E'+calculate+the+Energy+in+terms+of+E+at+maximum+height%3F&hl=en&gl=in&pid=bl&srcid=ADGEESi5cBCQKAef6Ik7dIWImfOEFmcDPd1rmTmX8A-P-yc8ENnh2-jsFd2pkwrDbYeYYQ6tjZNAG5Jfbn26aeZBkmCblk5NRBKTu2oiWM7vicllMxz9WpcqwaQ1gMCjMJ4UkwN8vo_A&sig=AHIEtbT7NozkVlaikcxMBbgMOdWDkWGBzg

Q-3

Answer 26

I would instinctively say: KEtop = KE horizontal = KE0 cos²30