## khanitha Group Title A piece of marble is projected from earth's surface with velocity of 50 m/s. 2 seconds later, it just clears a wall 5 m high. What is the angle of projection 2 years ago 2 years ago

1. Awstinf Group Title

Maybe you could make a triangle out of it so 2 seconds it barely clears the wall so the hyp is 100m and the opposite side from the angle is 5m and use inverse sin to find the angle so oppo/hyp 5/100, I got 2.87 degrees that seems a little small though...

2. Awstinf Group Title

|dw:1333208034298:dw| heres a picture

3. Awstinf Group Title

apparently its 60 degrees says another site

4. salini Group Title

@awstinf u totally ignord gravity acting downward(it gives the body accelerated motion)not uniform

5. salini Group Title

so wat u do is in horizontal direction,the cmponent of velocity is 50cos@ in time 2 seconds it will cover a distance 50cos@*2 |dw:1333208698171:dw| tan@=sin@/cos@=100/50cos@ solve it

6. amistre64 Group Title

shouldnt there be a distance part to this?

7. salini Group Title

|dw:1333208865379:dw| sry i made a mistake in the diagram

8. salini Group Title

?

9. amistre64 Group Title

lol, i spose we are doing: 2(50) cos(a) to determine a distance from the wall :)

10. amistre64 Group Title

$h(t)=-\frac{g}{2}t^2+V_o sin(a)\ t+H_o$ $h(2)=5=-9.8+2(50)sin(a)+0$ $5+9.8=100sin(a)$ $\frac{14.8}{100}=sin(a)=.1480$ $a=sin^{-1}(.1480)$ maybe

11. salini Group Title

something has caught me! i am reading all the question wrong1 |dw:1333209509913:dw| use 5m instead of 100m that gives tan@=sin@/cos@=5/100cos@ sin@=1/20=0.05

12. salini Group Title

amistre ur approach is absolutely right check ur 9.8t^2 part u gobbled up the t^2

13. amistre64 Group Title

:) yeah, i seen something was amiss, but couldnt place it. thanx

14. hosein Group Title

Hi use this formula$y=(-1/2)g t^2+v _{0}\sin \theta+y _{0}$ if you take y=5,g=9.81,t=2,v0=50,y0=0 theta yields29.49 degrees