Question

A ball is thrown Into air with a speed of 62 ms^-1 at an angle of 45° with the horizontal. Calculate 

1. the maximum height attained 

2. the horizontal range 

3. the time of flight and 

4. the velocity of the ball after 4 seconds.

Answer 1

1. Maximum height, H = \(\frac{u^2sin^2 \theta}{2g}\)Here,

u = 62 ms^-1 , θ = 45°, g = 9.8 ms^-2

⁼ \(\frac {62 \times 62}{4 \times 9.8}\) = 98.06 m.

2. Horizontal range. R is given by,

R = \(\frac{u^2sin2 \theta}{2g}\)

= \(\frac{ 62 \times 62 \times sin \,90}{9.8}\)

= \(\frac {62 \times 62}{4 \times 9.8}\) = 392.2m.

3. The time of flight,

T = \(\frac{2usin\theta}{g}\)

= \(\frac{ 2 \times 62 \times sin \,45}{9.8}\)

= \(\frac {2 \times 62}{\sqrt2 \times 9.8}\) = 8.95 s.

4. The horizontal component of the velocity remains constant throughout.

∴ Horizontal component of the velocity after 4 seconds v_x = ucosθ Here u = 62 ms^-1 and θ = 45°

∴ v_x = 62 cos45° = \(\frac{ 62}{\sqrt2}\) = 44.3 s.

The vertical component of the velocity changes due to the acceleration due to gravity.

We have v_y = u_y + at,

Here, u_y = usinθ; a = -g; t = 4 s;

From v_y = usinθ - 4g

∴ v_y = 62 sin45 - 9.8 × 4

\(\frac{ 62}{\sqrt2}\) -39.2 = 4.64 ms^-1

Magnitude of the resultant velocity after 4 seconds is,

V = \(\sqrt {v_x^2+v_y^2}\)

\(\sqrt {(44.3)^2+(4.64)^2} \)

= 44.5 ms^-1

Direction of the velocity is given by

tan a = \(\frac {v_y}{v_x}\) = \(\frac {4.64}{44.3} = 0.104 \)

∴ a = tan^-1 (0.104) = 5°5'