Question

a body is projected with velocity 60 m/s at an angle of 30°with horizontal. calculate maximum height, time of flight, horizontal range of Projectile if g = 10 m/s ²​

Answer 1

Answer:hmax=45m

Range=180root3

Time of flight=6 sec

Explanation:

Answer 2

The maximum height of the projectile is 45m, time of flight is 6s and the horizontal range is 180√3m

Step-by-step Explanation:

Given: Velocity = 60 m/s

Angle with horizontal = 30°

Acceleration due to gravity g = 10 m/s

To Find: Maximum height, time of flight, the horizontal range of projectile

Solution:

• Finding the maximum height

The maximum height reached by a projectile is,

$H = \frac{u^{2} sin^{2} \theta}{2g}$

Substituting the given values, we get

$H = \frac{(60)^{2} sin^{2} 30^o}{2 \times 10} = 45 m$

• Finding the time of flight

The time of flight of a projectile is the sum of time of ascent and time of descent. Therefore, we have

$t = t_{ascent} + t_{descent} = \frac{2u sin \theta}{g}$

Substituting the given values, we get

$t = \frac{2 \times 60 \times sin 30^o}{10} = 6 s$

• Finding the horizontal range

The horizontal range of a projectile is,

$R = \frac{u^{2} sin2 \theta}{g}$

Substituting the given values, we get

$R = \frac{(60)^{2} sin(2 \times 30^o)}{\times 10} = 180\sqrt{3} \ m$

Hence, The maximum height of the projectile is 45m, time of flight is 6s and the horizontal range is 180√3m