Question

A ball is kicked at an angle of 35° with the ground.
a) What should be the initial velocity of the ball so that it hits a target that is 30 meters away at a height of 1.8 meters?
b) What is the time for the ball to reach the target?

Answer 1

Answer:

hola mate

here is ur answer

a) 

x = V0 cos(35°) t 

30 = V0 cos(35°) t 

t = 30 / V0 cos(35°) 

1.8 = -(1/2) 9.8 (30 / V0 cos(35°))2 + V0 sin(35°)(30 / V0cos(35°)) 

V0 cos(35°) = 30 √ [ 9.8 / 2(30 tan(35°)-1.8) ] 

V0 = 18.3 m/s 

b) 

t = x / V0 cos(35°) = 2.0 s

tq♥️

Answer 2

Given:

Angle of throwing of projectile $= 35$

Horizontal Range of the projectile $= 30 m$

Maximum height reached by the projectile $= 1.8 m$

To find:

1. Initial velocity of throwing of the projectile (u).
2. Time taken to reach the target (t).

Solution:

Step 1

We have been given that the projectile is launched at some initial velocity (lets say "u") from the starting position at an angle of 35°.

Hence,

The initial velocity could be resolved into horizontal and vertical components namely $u .cos35$ in horizontal direction and $u .sin35$ in vertical direction.

Also, we know, during the path followed by projectile, the vertical component of the velocity becomes 0 at certain point whereas the horizontal component of velocity remains constant throughout  and acts as a motion in 1D.

Therefore,

$u.cos35=\frac{30}{t}$  (speed = distance traveled /time taken)

Hence, $t=\frac{30}{u.cos(35)}$      $(i)$

Now,

We have been given the horizontal range $= 30 m$

and height $= 1.8 m$

Step 2

Now, for the motion of launch projectile in the upward direction,

We have,

$h = 1.8 m$

$u = u .sin(35)$  (velocity for the upwards motion)

$t=\frac{30}{u.cos(35)}$

$a = -g = -10 m/s^{2}$

Hence, substituting the given values in the equation, we get

$h = ut-\frac{1}{2}gt^{2}$

$1.8=u. sin35(\frac{30}{u. cos(35)} )-\frac{1}{2}(10)(\frac{30}{u .cos(35)} )^{2}$

Therefore,

$1.8=30.tan(35)-\frac{4500}{u^{2}.cos^{2}(35) }$

We know, $tan(35)=0.7$  $and$  $cos(35)=0.8$

$1.8=30(0.7)-\frac{4500}{u^{2}(0.8)^{2} }$

$\frac{4500}{u^{2}(0.64) } = 21-1.8$

$\frac{7031}{u^{2} } = 19.2$   $;$  $or$

$u^{2}=366$    $;$  $hence$

$u = 18.6 m/s$

Step 3

Now,

Substituting this value in (i) ,we get

$t=\frac{30}{u. cos(35)}$

$t=\frac{30}{18.6(0.8)}$

Therefore,

$t=2.02s$

Final answer:

Hence,

1. The initial velocity of the ball is 18. m/s.
2. The ball reaches the target in 2.02 seconds.