Question

A body is projected with a velocity of 49m/s at angle 60 degrees with horizontal .It reaches a maximum height of

Answer 1

hope this help you mate...

Answer 2

Answer:

The maximum height reached by the projectile is 91.875 meter.

Explanation:

Initial velocity of projectile (u) = 49 m/s

Angle of projection (θ) = 60°

Let, the maximum height achived

by projectile (h) = H m

Gravitational acceleration (g) = 9.8 m/sec^2

From Projectile Motion in Kinematics we know;

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

From equation (i) we can calculate the maximum height;

$H = \frac{ {u}^{2} {\sin}^{2} \theta }{2g} \\ or \: H =\frac{ {(49)}^{2} {\sin}^{2} (60 \degree) }{2 \times 9.8} \: m \\ = 91.875 \: meter$

Hence we get the maximum height reached by the projectile is 91.875 meter .