Question

9. The maximum horizontal range of a projectile is given by the formula R= u2/g where u is the initial velocity and g is the acceleration due to gravity. Find the velocity with which a ball can be thrown to have a maximum range of 20 meters when the acceleration due to gravity is equal to 9.8 m/s.
(SHOW WORK)

Answer 1

Answer:

The velocity with which a ball must be thrown to have a maximum range of 20 m is 14 m/s.

Note that this problem means to find the magnitude of the velocity and not the direction (it is implicit in the formula that the angle of the launch is 45°).

Explanation:

You just must use the given equation for the maximum horizontal range of a projectile and solve for u which is the unknwon:

• Given equation: R = u² /g

• g = 9.8 m/s²

• R = 20 m

• u =?

Solve for u:

• u² = R × g = (20 m) × (9.8 m/s²)  =  196 m²/s²

Take square root from both sides:

• u = 14 m/s ← answer

Answer 2

Hello!

The answer is:

The velocity with which the ball can be thrown to have a maximum range of 20 meters is equal to 14 m/s.

$u=14\frac{m}{s}$

Why?

To solve the problem and find the velocity, we need to isolate it from the equation used to calculate the maximum horizontal range.

We have the equation:

$R=\frac{u^{2} }{g}$

Where,

R is the maximum horizontal range.

u is the initial velocity.

g is the gravity acceleration.

Also, from the statement we know that:

$R=20m\\g=9.8\frac{m}{s^{2} }$

So, using the given information, and isolating, we have:

$R=\frac{u^{2} }{g}$

$R*g=u^{2}$

$u^{2}=R*g=20m*9.8\frac{m}{s^{2} }=196\frac{m^{2} }{s^{2} }\\\\u=\sqrt{196\frac{m^{2} }{s^{2}}}=14\frac{m}{s}$

Hence, we have that the velocity with which the ball can be thrown to have a maximum range of 20 meters is equal to 14 m/s.

$u=14\frac{m}{s}$

Have a nice day!