Question

You launch a projectile from level ground at a speed of 25.0 m/sm/s and an angle of 36.9∘∘ above the horizontal. How long after it is launched does the projectile reach its maximum height above the ground?

Answer 1

Answer:

The projectile takes 2.04 s to reach its maximum height.

Explanation:

The maximum height of the projectile is given by the relation :

$H=\frac{u^{2}\sin^{2}\theta }{2g}$      .....(1)

Here H is height, u is initial speed, g is acceleration due to gravity and θ is projectile angle with respect to horizontal.

Substitute 25 m/s for u, 9.8 m/s² for g and 36.9° for θ in equation (1).

$H=\frac{(25)^{2}\times\sin^{2}\36.9 }{2\times9.8}$

H = 11.50 m

The equation of motion for vertical direction is :

v = u cosθ - gt

Here v is the vertical speed and t is time.

At maximum height, v is zero. So, the above equation in terms of t becomes :

t = (u cosθ)/g

Substitute the suitable values in the above equation:

t = $\frac{25\times\cos36.9}{9.8}$

t = 2.04 s