Question

A stone is thrown straight up from a cliff 20 meters above the ground with an initial velocity of 49 meters per second. Because gravity is the force pulling the stone down, the initial acceleration of the arrow is-9.8 meters per second squared. The equation that represents this relationship is H = - 9.8t ^ 2 + 49t + 20 where H is the height in meters and it is the time in seconds that the stone is in the air.

a. What is the maximum height that the stone reached?

b. How long would it take for the stone to reach the maximum height?

c. How many seconds it would take for the stone to hit the ground?

d. How high was the stone 2 seconds after it was thrown?​

Answer 1

Answer:

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Answer 2

Given:

Height of the cliff $=20m$

Initial velocity $(u)=49m/s$

Equation for height $H=-9.8t^{2}+49t+20$

To find:

1. Maximum height reached by the ball.
2. Time taken by the ball to reach the maximum height.
3. Time taken to reach the ground.
4. Distance of stone after 2 s of throw of ball.

Solution:

Step 1

We know, the ball is thrown from the top of a cliff of height 20 m with an initial velocity of 49 m/s.

The maximum height reached by the ball will be given by,

$v^{2}- u^{2}=2as$

We have,   $v=0m/s$  ;  $u=49m/s$  ; $a=-g=-9.8m/s^{2}$

Hence,

$(0)^{2}- (49)^{2}=2(-9.8)h$

$h=122.5m$

Hence, if a ball is thrown with a velocity of 49 m/s, the ball will reach a height of 122.5 m from the top of the cliff.

Step 2

We have now, that the stone reaches 122.5 m when thrown with an initial velocity of 49 m/s.

Hence, the time taken to reach the height will be given by

$H=-9.8t^{2}+49t+20$

We have,  $H=122.5m$   ;

$122.5=-9.8t^{2}+49t+20$

$9.8t^{2}-49t+120.5=0$

Solving the equation, we get

$t=2.5s$  $(approx)$

Step 3

Now,

When the ball after completing the projectile, falls on the ground, then, on the ground, the height of the ball from the ground is definitely zero.

Hence, applying this information in the equation we have, we get

$H=-9.8t^{2}+49t+20$

Here, $H=0$

$-9.8t^{2}+49t+20=0$

$9.8t^{2}-49t-20=0$

Solving this equation, we get

$t=5.3s$ $(approx)$

Step 4

We know, the ball takes 2.5 s to reach a height of 122.5 m. Hence if we substitute $t=2s$ in the given equation $H=-9.8t^{2}+49t+20$, we get the height of the ball after 2 s.

Hence,

$H=-9.8(2)^{2}+49(2)+20$

$H=-39.2+98+20$

$H=78.8m$

Hence, after 2 s, the ball will be at a height of 78.8 m.