Question

Q.75. A body is thrown with a velocity of 20 m/s at an angle of 45° with the horizontal.
a) What is the time taken by the body to reach the ground?
b) What is the maximum height reached by the body?
c) How far from the starting point does it reach the ground?
Take g = 10 m/s2. (2.82m; 10 m; 40 m)

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Answer 1

Answer:

(a) 2√2 seconds

(b) 10m

(c) 40m

Explanation:

The body is experiencing projectile motion.

So, Time of flight = 2u Sinθ / g

g is given as 10m/s².

u = 20m/s²

Angle θ = 45°

(a) Total time taken = (2×20sin45)/10

= (40 × 1/√2)/10

= 40/√2 × 1/10 = 2√2 sec

Approximately 2.82 seconds

(b) Maximum Height

H = (u Sinθ)²/2g

[u sinθ = 20sin45 = 20/√2 = 10√2]

H = (10√2)²/2 × 10

= 200/20 = 10m

(c) The displacement from the initial point ( Range )

R = u cosθ t

= 20cos45 × 2√2

= 20/√2 × 2√2 = 40m