Question

What is the range of ball which projected with velocity of 29.4m/s. Just passes over pole of 4.9m high

Answer 1

Given,

Velocity of projected ball = 29.4m/s

Maximum height of the ball = 4.9m

To find,

Range of the ball

Solution,

Let us take gravitational acceleration(g) as 9.8m/s².

The Formula for Range = v²sin2θ/g

The Formula for Max. Height = v²sin²θ/2g

As we already know velocity and gravity, the only unknown factor for range is sin2θ.

Using the formula for Max. Height, we can do the following-

=> v²sin²θ/2g = 4.9

=> (29.4)²sin²θ/(2*9.8) = 4.9

By solving this we get, sin²θ=1/9 => cos²θ=8/9 , as sin²θ+cos²θ=1

Sinθ=1/3, Cosθ=(2√2)/3

As, sin2θ = 2sinθcosθ => sin2θ = 2*(1/3)*((2√2)/3) => sin2θ = (4√2)/9

Range = v²sin2θ/g

After putting in the respective values we get,

=> (29.4)²((4√2)/9)/9.8 = 39.2√2 = 55.43m

Hence, the range of the ball is 55.43m