Question

A pebble is thrown into the air with a velocity of 19/m at an angle of 36 with respect to the horizontal.

if air resistance is negligible, what is the maximum height the pebble reaches?
A. 6.4 m
B. 8.0 m
C. 11 m
D. 19 m

Answer 1

The maximum height the pebble reaches is approximately;

A. 6.4 m

Step-by-step explanation:

The question is with regards to projectile motion of an object

The given parameters are;

The initial velocity of the pebble, u = 19 m/s

The angle the projectile path of the pebble makes with the horizontal, θ = 36°

The maximum height of a projectile,
`h_(max)`, is given by the following equation;

`h_(max) = (\left (u * sin(\theta) \right)^2)/(2 \cdot g)`

Therefore, substituting the known values for the pebble, we have;

`h_(max) = (\left (19 * sin(36 ^(\circ)) \right)^2)/(2 * 9.8) = 6.3633894140470403035477570509439`

Therefore, the maximum height of the pebble projectile,
`h_(max)` ≈ 6.4 m.