Question

If a ball is thrown at 30° angle Direc. from (20m) tower with intial speed of 30 m per sec (i) find the time taken by the ball to reach the ground.​

Answer 1

Answer:

TIME-2.473s

35.53°

Explanation:

A ball is thrown 30 degrees above the horizontal at 20m/s form a height of 30m above the ground. What is its velocity on impact with the ground below?

First, we need to find the total flight time, t. To do this,

Let’s review the 4 fundamental kinematic equations of motion for constant acceleration (strongly recommend you commit these to memory – they will serve you well):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

The total flight time of the ball will be the time taken to drop the 30m to the ground. This means that we use equation (1) to get t

We know that u = 0 (no velocity in the downward direction), s = 30m, a = g, and we want to find t:

s = ut + ½at^2

30 = 0 + 4.905t^2

So t = √(30/4.905) = 2.473s

Now we can use the equations for projectile motion:

At any time t, a projectile's horizontal and vertical displacement are:

x = VtCos θ where V is the initial velocity, θ is the launch angle

y = VtSinθ – ½gt^2

The velocities are the time derivatives of displacement:

Vx = VCosθ (note that Vx does not depend on t, so Vx is constant)

Vy = VSinθ – gt

So at time t = 2.473s,

Vx = 20Cos30° = 17.321m/s

Vy = 20Sin30° - 9.81(2.473) = 24.260m/s

So the total velocity on impact is √(17.321^2 + 24.260^2) = 29.81m/s at an angle of Tan^-1(17.321/24.26) = 35.53°