Question

(d) Suppose a body is projected at angle 45 with the horizontal and the magnitude of the velocity is 400 ms. If the initial height is 3 m, determine the maximum height reached and the time to reach ground level. Laninantal

Answer 1

The initial height is 3 m.

the = 3 m

The initial vertical velocity is

4, = 400 sin 45°

And the acceleration in the vertical direction is

ay = -9= -9.8 m/s2

Now, we know that at the maximum height the vertical velocity of the projectile is zero. So, the maximum height can be given by

Hmar = h. + 2ahot

Putting the values

Hmar = 3+ (400 sin 45°) L=4081.6 m 2 x 9,8

So, the maximum height is 4081.6 m.

Now, the total final vertical displacement is -3 m (why? because finally it get to the ground which is 3 m down the initial height). i.e

d= -3 m

We know

d = ut + -ať = uyt - =gt?

9.8 →-3= (400 sin 45°) -

4.96 - 282.84t - 3=0

The positive solution to this quadratic equation is

t = 57.73

So, the time to reach the ground is 57.73 s.