1. ## Physics Question

1) A missile is shot from the ground with an initial speed of 2,500 m/s at an angle of 40 degrees. Find how high the missile reaches, the time in the air, and horizontal range.

2) If the missile was shot at a platform 3 meters above the ground, how high would the missile reach, it's time in the air? and horizontal range?

(Neglect Air Resistance)
---

Velocity in the Y: Sin(40)*2500= 1606 m/s
Velocity in the X: Cos(40)*2500= 1915 m/s

Time to reach maximum height: 0=(1606)-(9.80)t= 164 seconds

Max height above cliff: (1/2)(9.8)(164)+(1606)(164)= 147846 meters

Time to fall from max height: 147846=(1/2)(9.8)t^2+0t
t^2= 2(147846)/9.8 = 30173
t= sqrt(30173)= 174 seconds

Total time in air= 174+164= 338 seconds

Horizontal Range: (1915)(338)= 647270
---

Anyway, if anyone could confirm or deny the validity of the problem above I'd appreciate it, and as for the second problem... I'm not sure how to start it.

2. Originally Posted by Caturdayz
1) A missile is shot from the ground with an initial speed of 2,500 m/s at an angle of 40 degrees. Find how high the missile reaches, the time in the air, and horizontal range.

2) If the missile was shot at a platform 3 meters above the ground, how high would the missile reach, it's time in the air? and horizontal range?

(Neglect Air Resistance)
---

Velocity in the Y: Sin(40)*2500= 1606 m/s
Velocity in the X: Cos(40)*2500= 1915 m/s

Time to reach maximum height: 0=(1606)-(9.80)t= 164 seconds

Max height above cliff: (1/2)(9.8)(164)+(1606)(164)= 147846 meters

Time to fall from max height: 147846=(1/2)(9.8)t^2+0t
t^2= 2(147846)/9.8 = 30173
t= sqrt(30173)= 174 seconds

Total time in air= 174+164= 338 seconds

Horizontal Range: (1915)(338)= 647270
---

Anyway, if anyone could confirm or deny the validity of the problem above I'd appreciate it, and as for the second problem... I'm not sure how to start it.
HI

For (1) ,

$v^2=u^2+2as$

$0=(2500\sin 40)^2-2(9.8)(H)$

Greatest height , $H = 131752.5$ m

Here , i am not sure what time in the air means .. is it time taken to reach maximum height , or time of flight .

Anyways , if its time taken to reach max height ,

$v=u+at$

$0=(2500\sin 40)-(9.8)t$

$t=164$ sec

Then time of flight would be 164 x 2 = 328 since air resistance is neglected .

Horizontal range , $s = vt$

$s=(2500\cos 40)(328)$

$=628156.4$ m

3. Originally Posted by Caturdayz
1) A missile is shot from the ground with an initial speed of 2,500 m/s at an angle of 40 degrees. Find how high the missile reaches, the time in the air, and horizontal range.

2) If the missile was shot at a platform 3 meters above the ground, how high would the missile reach, it's time in the air? and horizontal range?

(Neglect Air Resistance)
---

Velocity in the Y: Sin(40)*2500= 1606 m/s
Velocity in the X: Cos(40)*2500= 1915 m/s

Time to reach maximum height: 0=(1606)-(9.80)t= 164 seconds

Max height above cliff: (1/2)(9.8)(164)+(1606)(164)= 147846 meters

Time to fall from max height: 147846=(1/2)(9.8)t^2+0t
t^2= 2(147846)/9.8 = 30173
t= sqrt(30173)= 174 seconds

Total time in air= 174+164= 338 seconds

Horizontal Range: (1915)(338)= 647270
---

Anyway, if anyone could confirm or deny the validity of the problem above I'd appreciate it, and as for the second problem... I'm not sure how to start it.
For (2) , i assume it still uses the info in (1) .

$s=ut+\frac{1}{2}at^2$

$-3=(2500\sin 40)t-\frac{1}{2}(9.8)t^2$

Displacement is negative , then take the non zero value of t .

Horizontal range , $s =(2500\cos 40)(t)$

The t you calculated above .

For greatest height , you are now just 3 m above where you just fired just now so what would happen to the height of the missle ? neglecting air resistance .