A mortar shell is fired with horizontal velocity of 300 m/s form the top of cliff 100 m high. (i) How long will it take to reach the ground ? (ii) How far from the foot of cliff will it strike the ground ?

Question

king · Answer

time=60 s?

king · Answer

18000 m?

king · Answer

is it rite?

king · Answer

hasnain?

anonymous · Answer

ooh sorry , its is wrong , i have done mistake sorry , i am writting it now

anonymous · Answer

A mortar shell is fired at a ground level target 490m away with initial velocity of  98m/s. Find the two possible values of the launch angle. Calculate the minimum time to hit the target.

amistre64 · Answer

assuming an unreal parabolic path .... we need to calculate this equation; which turns into the physics equation for it in the end that I can never remember :)

$$-\frac{1}{2}g\ t^2+V_i sin(a)\ t + H_i=0$$
$$-\frac{1}{2}9.8\ t^2+98 sin(a)\ t + 0=0$$

we should know that the outward speed is: 98 cos(a); and that it needs to cover a distance of 480 in t sec.

98 cos(a) t = 480

t = $\cfrac{480}{98cos(a)}$

lets use this "value" for t in out equation:

$$-\frac{1}{2}9.8\ \left(\cfrac{480}{98cos(a)}\right)^2+98 sin(\theta)\ \cfrac{480}{98cos(a)}=0$$
$$-\frac{480^2(4.9)}{98^2}sec^2(a)+ 480\ tan(a)=0$$
$$-\frac{480^2(4.9)}{98^2}(tan^2(a)+1)+ 480\ tan(a)=0$$
$$-\frac{480^2(4.9)}{98^2}tan^2(a)+ 480\ tan(a)-\frac{480^2(4.9)}{98^2}=0$$

this of course is a quadratic equation that can be solved for tan(a)

amistre64 · Answer

$$tan(a)=\frac{-480\pm \sqrt{480^2-\cfrac{4(480^4)(4.9^2)}{98^4}}}{\cfrac{-2(480^2)(4.9)}{98^2}}$$

that looks fun lol ... better let the wolf handle it :)

tan(a) = .26166  and  3.82167

this means that:

  a = arctan(.26166) or arctan(3.82167)

  a = 14.66  or 75.34 degrees

but after all that youd have to dbl chk it

amistre64 · Answer

and of course I used 480 as my standard typo to mess it all up; ... gotta get to class anyhoos so good luck :)