# A projectile is fired at an angle of 43deg from horizontal. The intial speed is 30m/s. A 10meter high wall is placed 80m from the firing position. Will it get over the wall?

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### 1 Answer

The projectile is fired at an angle of 43 degrees with the horizontal at an initial speed 30 m/s.

The initial velocity of the projectile can be divided into two components, a horizontal component equal to 30*cos 43 and a vertical component 30*sin 43. The acceleration of the projectile in the horizontal direction is 0, while it is 9.8 m/s^2 in the vertically downwards direction. If the time taken by the projectile to travel 80 m in the horizontal direction is t, 30*cos 43*t = 80

=> t = `80/(30*cos 43)`

The height of the projectile at time t = `80/(30*cos 43)` is equal to 30*sin `43*80/(30*cos 43) - (1/2)*9.8*(80/(30*cos 43))^2`

= `tan 43*80 - (4.9*64)/(9*cos^2 43)`

= 9.45 m

The height of the wall 80 m away from where the projectile is fired is 10 m, but the height of the projectile is only 9.45 m.

**The projectile does not go over the wall.**