Question

The maximum horizontal distance a boy can throw a ball is 50 m. Assume he can throw with the same initial speed at all angles. How high does he throw the ball when he throws it straight upward?

a) 25 m
b) 12.5 m
c) 6.25 m
d) 50 m

Answer 1

Using the principles of projectile motion, we calculate that the maximum height the boy can throw the ball straight upwards is 12.5 m, based on the maximum range of 50 m at an optimal angle of 45 degrees.

Step-by-step explanation:

To determine how high the boy can throw the ball when he throws it straight upward, we use the principles of projectile motion. Considering that the maximum range (50 m) is achieved at an angle of 45 degrees for a projectile without air resistance, the horizontal and vertical components of the initial velocity are equal.

Using the equation for vertical motion (y = vy2 / (2g)), where vy is the vertical component of the initial velocity and g is the acceleration due to gravity, we can find the maximum height.

Given that the range R = (vi2 sin 2θ) / g, and at 45 degrees, sin2θ = 1, we can deduce that vi2 = Rg. So the vertical component of the initial velocity vy would be sqrt(Rg)/sqrt(2), and we can say: y = (Rg)/(4g). Plugging in R = 50 m and g = 9.8 m/s2, we get the maximum height, y.

Thus, the boy can throw the ball to a maximum height of 12.5 m when throwing it straight upwards.