# Physics Question

• Oct 11th 2009, 01:28 PM
Caturdayz
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.
• Oct 17th 2009, 11:24 PM
Quote:

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
• Oct 17th 2009, 11:36 PM
Quote:

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 .