Question

32.
A ball is thrown at angle with the horizontal & another ball is thrown at an angle (90 - 0) with
the horizontal from the same point on horizontal ground with the same speed of 39.2 m/s. The
second ball attains a maximum height of 50m more as compared to the maximum height attained by
the first ball. Find the maximum height, attained by each of the ball. (g=9.8 m/s²)​

Answer 1

Answer:

24.5 m  & 74.5 m

Explanation:

Let say Velocity of A = 39.2 at angle α

Velocity of B = 39.2 at angle (90-α)

Vertical Velocity of A = 39.2Sinα

At max height Vertical velocity = 0

using V² - U² = 2aS

a = -g = -9.8 m/s²

=> S = -(39.2)²Sin²α/(2 * (-9.8))

=> S = 78.4Sinα

Similalrly for B

Vertical Velocity of A = 39.2Sin(90-α) =  39.2Cosα

S = 78.4Cosα

78.4Sinα - 78.4Cosα  = 50

=> Sinα - Cosα = 50/78.4

Squaring both sides

=> Sin²α + Cos²α -2SinαCosα = (50/78.4)²

=> 1 - Sin2α = (50/78.4)²

=> Sin2α = 36°   (approx)

=> α = 18°

78.4Sinα = 78.4Sin18 = 24.5 m    

78.4Cosα = 74.5 m

the maximum height, attained by each of the ball = 24.5 m  & 74.5 m