Question

An object is thrown with initial speed 5m/s with an angle projection of 30° . What is the range and hieght reached by the particle ?

Answer 1

Initial velocity, u = 5m/s
Angle, x = 30°

Maximum height reached:
h = (u sinx)^2/2g

h = (5×sin 30)^2/2×9.8

h = (5×0.5)^2/2×9.8

h = 0.32 m


Range:
R = u^2 sin2x/g

R = 5^2 × sin 60 / 9.8

R = 25×√3/ (2×9.8)

R = 2.21 m


Range is 2.21m and maximum height reached is 0.32m

Answer 2

Hope u like my process
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=> Initial velocity (u) = 5 m/s

=> Angle of projection = 30°

=> Height attained by object

$= > \bf \: \frac{ {(u \sin( \theta ) )}^{2} }{2g} = \frac{ {(5 \times \frac{1}{2}) }^{2} }{2 \times 9.8} \\ \\ \bf \: = \frac{25}{4 \times2 \times 9.8} = \underline{0.3189} \: m {s}^{ - 1}$

=> Range of the projectile

$= > \bf \frac{ {u}^{2} \sin(2 \theta ) }{g} = \frac{ {5}^{2} \times \sin(60) }{9.8} \\ \\ = \bf \frac{25 \times \sqrt{3} }{2 \times 9.8} = \underline{2.209} \: \: m$
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