Question

A body is projected with an initial velocity of 20 ms -1

at an angle of 30° with the horizontal. Calculate

(a) maximum height, (b) time taken to reach the maximum height and (c) horizontal range.​

Answer 1

Explanation:

For Max Height u Can use ...H max =

20×20×1/2×10×4

= 5m ( g is taken 10 here )

Time taken = 2Usinbtehta/g

2×20×1/10×2

= 2 sec

Range = U2 Sin 2tehta /g

= 20× 20 × ✓3/2×10

= 20✓3 m (g is taken 10 here)

Answer 2

Answer:

Use formula of

Max. height =U^2 sin^2theta/2g

Time taken =2 U Sin theta /g

Horizenta range =U^2 Sin2 theta /g

Explanation:

where g is 10m/s^2