Question

A projectile was thrown up at 60 degree with the horizontal at an initial speed of 5.0 m/s. Find the maximum height, time of flight, and range of the projectile

Answer 1

Answer:

Explanation:

given :

        θ = 60°

        u = 5.0 m/s       where u  ⇒ initial horizontal speed.

to find:

        1) maximum height (h)

        2) time of flight (T)

        3) range (R)

we can use the below formula to find the values.

h = ($u^{2}$ $sin^{2}$θ) / 2 g

T = (2 u sinθ) / g

R = ($u^{2}$ sin2θ) / g

now  h = (25 × (3/4)) / 19.6           since  $sin^{2}$ 60 = 3/4   and   g= 9.8m/$s^{2}$

        h=  0.9567 m

and  T = (10 × √3/2) / 9.8             since sin 60 = √3/2

        T =  0.8837 seconds

and last ,

R = (25×√3/2) / 9.8           since   sin 120 = √3/2

R = 2.209 m