Question

A ball is thrown at an angle of 45 degrees to the ground, and lands 25 meters away. the initial speed of the ball was

Answer 1

R=(v²×sin2theta)/g
R = v²/g (sin2theta = sin90° =1)
R= v²/g
v² = R×g = 25× 10=250
v= √ 250 = 15.8 m/s

Answer 2

The initial velocity of the ball is 15.6 m/s

GIVEN

A ball is thrown at an angle of 45 degrees to the ground and lands 25 meters away.

TO FIND

The initial speed of the ball.

SOLUTION

We can simply solve the above problem as follows;

To calculate the initial velocity of the projectile, we will apply the following formula;

$R = \frac{ {u}^{2} \sin2(θ) }{g}$

Where,

R = Range of the projectile = 25 m

g = Acceleration due to gravity = 9.8 m/s²

θ = angle of the projectile = 45°

Putting the values in the above formula we get;

$25 = \frac{ {u}^{2} \sin2(45) }{9.8}$

$u^{2} \sin(90) = 25 \times 9.8$

u² = 245

u = √245

u = 15.6 m/s

Hence, The initial velocity of the ball is 15.6 m/s

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