# 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?

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.

Approved by eNotes Editorial Team