Physics, asked by sanjeevsony7793, 8 months ago

A ball is thrown horizontally from a point 100m above the ground with a speed of 20m//s. Find (a) the time it takes to reach the ground, (b) the horizontal distance it travels before reaching the ground, (c) the velocity (direction and magnitude) with which it strikes the ground.

Answers

Answered by rahul123437

Given:

A ball is thrown horizontally from a point 100m above the ground with a speed of 20m/s. So $U_x$ = 20 m/s and $U_y$ = o m/s

To find:

(a) the time it takes to reach the ground.

(b) the horizontal distance it travels before reaching the ground

Formula used:

v= u+at

s= ut + $\frac{1}{2}$ at²

v² - u² = 2as

Explanation:

It is case of projectile fired horizontally from a height h = 100m

g = acceleration due to gravity = -9.81 m/s

a) time taken to reach the ground .

s= $U_y$ t + $\frac{1}{2}$ at² $U_y$ = 0m/s

t= $\sqrt{\frac{2h}{g} }$ = $\sqrt{\frac{2\times 100}{9.81} }$ = 4.51 sec.

b)The horizontal distance it travels before reaching the ground.

s= ut + $\frac{1}{2}$ at² Horizontal acceleration is zero

So horizontal distance = ut = 20 $\times$ 4.52 = 90.2 m

We can find the final velocity in x direction and y direction.

$V_x$ = $U_x$ +at Horizontal acceleration is zero

So, $V_x$ = $U_x$ = 20 m/s

Velocity in y direction.

$V_y$ = $U_y$ + $a_y$ t $U_y$ = 0m/s

$V_y$ = $a_y$ t = -9.8 × 4.51 = -44.19 m/s

Resultant velocity = $\sqrt{{v_x}^2 +{v_y}^2}$ = 48.4 m/s

Direction, ⇒ $\tan ^{-1} \frac{v_y}{v_x}$ = $\tan ^{-1} \frac{44.19}{20}$ = 66°

The ball strikes the ground with a velocity of 48.42 m/s at an agle 66° with horizontal.

To learn more....

1)Can we use the equation of kinematics to find the height attained by a body projected upward with any velocity

https://brainly.in/question/5157211

Previous Question

Next Question