Question

A projectile is thrown with an initial speed v₀ at an angle θ =24⁰ as shown in the figure. At time t_A the projectile reaches the point A with coordinates (d,h) where d=33m and h=5.3m in the given coordinate system.

Answer 1

In order to find the time it reaches point A, we need to use newton's equations.

The vertical component of initial velocity = u sin(theta)

The horizontal component of initial velocity = u cos(theta)

After any time t, the horizontal displacement is given by

x = u cos(theta) * t

33 = u cos(24) * t = 0.91 u*t ------------(1)

After any time t, the vertical displacement is given by

y = u sin(theta) * t - 1/2 * g *t2

5.3 = u sin(24) t - 0.5*9.8*t2

5.3 = 0.41 u*t - 0.49 * t2 -------------(2)

Solving 1 and 2, we get

t = 4.38 sec

u = 8.25 m/s